Expansions

We know that,

⇒ (a + b)² = a² + b² + 2ab.

(i) Given,

⇒ (4a + 9)²

⇒ (4a)² + (9)² + 2 × 4a × 9

⇒ 16a² + 81 + 72a.

Hence, (4a + 9)² = 16a² + 81 + 72a.

(ii) Given,

⇒ (3x + 10y)²

⇒ (3x)² + (10y)² + 2 × 3x × 10y

⇒ 9x² + 100y² + 60xy.

Hence, (3x + 10y)² = 9x² + 100y² + 60xy.

(iii) Given,

$$\begin{aligned}\Rightarrow (\sqrt{2}m + \sqrt{3}n)^2 \\[1em] \Rightarrow (\sqrt{2}m)^2 + (\sqrt{3}n)^2 + 2 \times \sqrt{2}m \times \sqrt{3}n \\[1em] \Rightarrow 2m^2 + 3n^2 + 2\sqrt{6}mn.\end{aligned}$$

Hence, $(\sqrt{2}m + \sqrt{3}n)^2 = 2m^2 + 3n^2 + 2\sqrt{6}mn$.

We know that,

⇒ (a + b)² = a² + b² + 2ab.

(i) Given,

⇒ (2a² + 3b)²

⇒ (2a²)² + (3b)² + 2 × 2a² × 3b

⇒ 4a⁴ + 9b² + 12a²b

Hence, (2a² + 3b)² = 4a⁴ + 9b² + 12a²b .

(ii) Given,

⇒ (3x²y + z)²

⇒ (3x²y)² + (z)² + 2 × 3x²y × z

⇒ 9x⁴y² + z² + 6x²yz

Hence, (3x²y + z)² = 9x⁴y² + z² + 6x²yz .

(iii) Given,

$$\begin{aligned}\Rightarrow \Big(2x + \frac{1}{3x}\Big)^2 \\[1em] \Rightarrow (2x)^2 + \Big(\frac{1}{3x}\Big)^2 + 2 \times 2x \times \frac{1}{3x} \\[1em] \Rightarrow 4x^2 + \frac{1}{9x^2} + \frac{4}{3}\end{aligned}$$

Hence, $\Big(2x + \frac{1}{3x}\Big)^2 = 4x^2 + \frac{1}{9x^2} + \frac{4}{3}$.

We know that,

⇒ (a + b)² = a² + b² + 2ab.

(i) Given,

$$\begin{aligned}\Rightarrow \Big(\frac{2}{5}x + \frac{5}{6}y\Big)^2 \\[1em] \Rightarrow \Big(\frac{2}{5}x\Big)^2 + \Big(\frac{5}{6}y\Big)^2 + 2 \times \frac{2}{5}x \times \frac{5}{6}y \\[1em] \Rightarrow \frac{4}{25}x^2 + \frac{25}{36}y^2 + \frac{4}{6}xy \\[1em] \Rightarrow \frac{4}{25}x^2 + \frac{25}{36}y^2 + \frac{2}{3}xy\end{aligned}$$

Hence, $\Big(\frac{2}{5}x + \frac{5}{6}y\Big)^2 = \frac{4}{25}x^2 + \frac{25}{36}y^2 + \frac{2}{3}xy$.

(ii) Given,

$$\begin{aligned}\Rightarrow \Big(\frac{x}{3} + \frac{6}{x}\Big)^2 \\[1em] \Rightarrow \Big(\frac{x}{3}\Big)^2 + \Big(\frac{6}{x}\Big)^2 + 2 \times \frac{x}{3} \times \frac{6}{x} \\[1em] \Rightarrow \frac{x^2}{9} + \frac{36}{x^2} + 4\end{aligned}$$

Hence, $\Big(\frac{x}{3} + \frac{6}{x}\Big)^2 = \frac{x^2}{9} + \frac{36}{x^2} + 4$.

(iii) Given,

$$\begin{aligned}\Rightarrow \Big(6 + \frac{5}{x}\Big)^2 \\[1em] \Rightarrow (6)^2 + \Big(\frac{5}{x}\Big)^2 + 2 \times 6 \times \frac{5}{x} \\[1em] \Rightarrow 36 + \frac{25}{x^2} + \frac{60}{x}\end{aligned}$$

Hence, $\Big(6 + \frac{5}{x}\Big)^2 = 36 + \frac{25}{x^2} + \frac{60}{x}$.

We know that,

⇒ (a - b)² = a² + b² - 2ab.

(i) Given,

⇒ (5x - 3y)²

⇒ (5x)² + (3y)² - 2 × 5x × 3y

⇒ 25x² + 9y² - 30xy

Hence, (5x - 3y)² = 25x² + 9y² - 30xy .

(ii) Given,

⇒ (3a - 7b)²

⇒ (3a)² + (7b)² - 2 × 3a × 7b

⇒ 9a² + 49b² - 42ab

Hence, (3a - 7b)² = 9a² + 49b² - 42ab.

(iii) Given,

$$\begin{aligned}\Rightarrow \Big(\frac{1}{2}x\ -\ \frac{3}{2}y\Big)^2 \\[1em] \Rightarrow \Big(\frac{1}{2}x\Big)^2 + \Big(\frac{3}{2}y\Big)^2 - 2 \times \frac{1}{2}x \times \frac{3}{2}y \\[1em] \Rightarrow \frac{x^2}{4} + \frac{9}{4}y^2 - \frac{3}{2}xy\end{aligned}$$

Hence, $\Big(\frac{1}{2}x - \frac{3}{2}y\Big)^2 = \frac{x^2}{4} + \frac{9}{4}y^2 - \frac{3}{2}xy$.

We know that,

⇒ (a - b)² = a² + b² - 2ab.

(i) Given,

$$\begin{aligned}\Rightarrow \Big(a^2 - \frac{b}{2}\Big)^2 \\[1em] \Rightarrow (a^2)^2 + \Big(\frac{b}{2}\Big)^2 - 2 \times a^2 \times \frac{b}{2} \\[1em] \Rightarrow a^4 + \frac{b^2}{4} - a^2b\end{aligned}$$

Hence, $\Big(a^2 - \frac{b}{2}\Big)^2 = a^4 + \frac{b^2}{4} - a^2b$.

(ii) Given,

$$\begin{aligned}\Rightarrow \Big(\frac{3a}{2b} - \frac{2b}{3a}\Big)^2 \\[1em] \Rightarrow \Big(\frac{3a}{2b}\Big)^2 + \Big(\frac{2b}{3a}\Big)^2 - 2 \times \frac{3a}{2b} \times \frac{2b}{3a} \\[1em] \Rightarrow \frac{9a^2}{4b^2} + \frac{4b^2}{9a^2} - 2\end{aligned}$$

Hence, $\Big(\frac{3a}{2b} - \frac{2b}{3a}\Big)^2 = \frac{9a^2}{4b^2} + \frac{4b^2}{9a^2} - 2$.

(iii) Given,

$$\begin{aligned}\Rightarrow \Big(5x - \frac{2}{3x}\Big)^2 \\[1em] \Rightarrow (5x)^2 + \Big(\frac{2}{3x}\Big)^2 - 2 \times 5x \times \frac{2}{3x} \\[1em] \Rightarrow 25x^2 + \frac{4}{9x^2} - \frac{20}{3}\end{aligned}$$

Hence, $\Big(5x - \frac{2}{3x}\Big)^2 = 25x^2 + \frac{4}{9x^2} - \frac{20}{3}$.

We know that,

⇒ (a + b + c)² = a² + b² + c² + 2(ab + bc + ca).

(i) Given,

⇒ (a + 2b + 3c)²

⇒ (a)² + (2b)² + (3c)² + 2 × (a × 2b + 2b × 3c + 3c × a)

⇒ a² + 4b² + 9c² + 2 × (2ab + 6bc + 3ca)

⇒ a² + 4b² + 9c² + 4ab + 12bc + 6ac

Hence, (a + 2b + 3c)² = a² + 4b² + 9c² + 4ab + 12bc + 6ac.

(ii) Given,

⇒ (3x + 5y - 2z)²

⇒ [3x + 5y + (-2z)]²

⇒ (3x)² + (5y)² + (-2z)² + 2 × [3x × 5y + 5y × (-2z) + (-2z) × 3x]

⇒ 9x² + 25y² + 4z² + 2 × (15xy - 10yz - 6xz)

⇒ 9x² + 25y² + 4z² + 30xy - 20yz - 12xz

Hence, (3x + 5y - 2z) = 9x² + 25y² + 4z² + 30xy - 20yz - 12xz.

(iii) Given,

⇒ (2x - 3y + 7z)²

⇒ [2x + (-3y) + 7z]²

⇒ (2x)² + (-3y)² + (7z)² + 2 × [2x × (-3y) + (-3y) × (7z) + 7z × 2x]

⇒ 4x² + 9y² + 49z² + 2 × [-6xy - 21yz + 14xz]

⇒ 4x² + 9y² + 49z² - 12xy - 42yz + 28xz.

Hence, (2x - 3y + 7z) = 4x² + 9y² + 49z² - 12xy - 42yz + 28xz.

We know that,

⇒ (a + b + c)² = a² + b² + c² + 2(ab + bc + ca).

(i) Given,

⇒ (6 - 2y + 4z)²

⇒ [6 + (-2y) + 4z]²

⇒ (6)² + (-2y)² + (4z)² + 2 × [6 × (-2y) + (-2y) × (4z) + 4z × 6]

⇒ 36 + 4y² + 16z² + 2 × [-12y - 8yz + 24z]

⇒ 36 + 4y² + 16z² - 24y - 16yz + 48z

Hence, (6 - 2y + 4z)² = 36 + 4y² + 16z² - 24y - 16yz + 48z.

(ii) Given,

⇒ (4x - 3y + z)²

⇒ [4x + (-3y) + z]²

⇒ (4x)² + (-3y)² + (z)² + 2 × [4x × (-3y) + (-3y) × (z) + z × 4x]

⇒ (16x)² + 9y² + z² + 2 × [-12xy - 3yz + 4xz]

⇒ 16x² + 9y² + z² - 24xy - 6yz + 8xz

Hence, (4x - 3y + z)² = 16x² + 9y² + z² - 24xy - 6yz + 8xz.

(iii) Given,

⇒ (7 - 2x - 3y)²

⇒ [7 + (-2x) + (-3y)]²

⇒ (7)² + (-2x)² + (-3y)² + 2 × [7 × (-2x) + (-2x) × (-3y) + (-3y) × 7]

⇒ 49 + 4x² + 9y² + 2 × (-14x + 6xy - 21y)

⇒ 49 + 4x² + 9y² - 28x + 12xy - 42y.

Hence, (7 - 2x - 3y)² = 49 + 4x² + 9y² - 28x + 12xy - 42y.

We know that,

⇒ (a + b + c)² = a² + b² + c² + 2(ab + bc + ca).

(i) Given,

$$\begin{aligned}\Rightarrow \Big(\frac{a}{2} + \frac{b}{3} + \frac{c}{4}\Big)^2 \\[1em] \Rightarrow \Big(\frac{a}{2}\Big)^2 +\Big(\frac{b}{3}\Big)^2 + \Big(\frac{c}{4}\Big)^2 + 2 × \Big[\big(\frac{a}{2}\big) × \big(\frac{b}{3}\big) + \big(\frac{b}{3}\big) × \big(\frac{c}{4}\big) + \big(\frac{c}{4}\big) × \big(\frac{a}{2}\big)\Big] \\[1em] \Rightarrow \frac{a^2}{4} + \frac{b^2}{9} + \frac{c^2}{16} + 2 × \Big[\big(\frac{ab}{6}\big) + \big(\frac{bc}{12}\big) + \big(\frac{ca}{8}\big)\Big] \\[1em] \Rightarrow \frac{a^2}{4} + \frac{b^2}{9} + \frac{c^2}{16} + \frac{ab}{3} + \frac{bc}{6} + \frac{ca}{4} \\[1em]\end{aligned}$$

Hence, $\Big(\frac{a}{2} + \frac{b}{3} + \frac{c}{4}\Big)^2 = \frac{a^2}{4} + \frac{b^2}{9} + \frac{c^2}{16} + \frac{ab}{3} + \frac{bc}{6} + \frac{ca}{4}$.

(ii) Given,

$$\begin{aligned}\Rightarrow \Big(\frac{2x}{3} + \frac{3}{2y} - 2\Big)^2 \\[1em] \Rightarrow \Big[\frac{2x}{3} + \frac{3}{2y} + (-2)\Big]^2 \\[1em] \Rightarrow \Big(\frac{2x}{3}\Big)^2 +\Big(\frac{3}{2y}\Big)^2 + (-2)^2 + 2 × \Big[\big(\frac{2x}{3}\big) × \big(\frac{3}{2y}\big) + \big(\frac{3}{2y}\big) × (-2) + (-2) × \big(\frac{2x}{3}\big)\Big] \\[1em] \Rightarrow \frac{4x^2}{9} + \frac{9}{4y^2} + 4 + 2 × \Big[\frac{6x}{6y} - \frac{6}{2y} - \frac{4x}{3}\Big] \\[1em] \Rightarrow \frac{4x^2}{9} + \frac{9}{4y^2} + 4 + 2 × \Big[\frac{x}{y} - \frac{3}{y} - \frac{4x}{3}\Big]\\[1em] \Rightarrow \frac{4x^2}{9} + \frac{9}{4y^2} + 4 + \frac{2x}{y} - \frac{6}{y} - \frac{8x}{3} \\[1em]\end{aligned}$$

Hence, $\Big(\frac{2x}{3} + \frac{3}{2y} - 2\Big)^2 = \frac{4x^2}{9} + \frac{9}{4y^2} + 4 + \frac{2x}{y} - \frac{6}{y} - \frac{8x}{3}$.

(iii) Given,

$$\begin{aligned}\Rightarrow \Big(2x + \frac{3}{x} - 1 \Big)^2 \\[1em] \Rightarrow \Big[2x + \frac{3}{x} + (-1) \Big]^2\\[1em] \Rightarrow (2x)^2 +\Big(\frac{3}{x}\Big)^2 + (-1)^2 + 2 × \Big[2x × \Big(\frac{3}{x}\Big) + \Big(\frac{3}{x}\Big) × (-1) + (-1) × (2x)\Big] \\[1em] \Rightarrow 4x^2 + \frac{9}{x^2} + 1 + 2 × \Big[6 - \frac{3}{x} - 2x\Big] \\[1em] \Rightarrow 4x^2 + \frac{9}{x^2} + 1 + 12 - \frac{6}{x} - 4x \\[1em] \Rightarrow 4x^2 + \frac{9}{x^2} + 13 - \frac{6}{x} - 4x \\[1em]\end{aligned}$$

Hence, $\Big(2x + \frac{3}{x} - 1 \Big)^2 = 4x^2 + \frac{9}{x^2} + 13 - \frac{6}{x} - 4x$.

(i) Given,

⇒ (x + 7)(x + 4)

⇒ x² + 4x + 7x + 28

⇒ x² + 11x + 28.

Hence, (x + 7)(x + 4) = x² + 11x + 28.

(ii) Given,

⇒ (a + 13)(a - 8)

⇒ a² - 8a + 13a - 104

⇒ a² - 5a - 104.

Hence, (a + 13)(a - 8) = a² + 5a - 104.

(iii) Given,

⇒ (y - 6)(y - 4)

⇒ y² - 4y - 6y + 24

⇒ y² - 10y + 24.

Hence, (y - 6)(y - 4) = y² - 10y + 24.

(i) Given,

⇒ (9 + 2x)(9 - 3x)

⇒ 81 - 27x + 18x - 6x²

⇒ 81 - 9x - 6x².

Hence, (9 + 2x)(9 - 3x) = 81 - 9x - 6x².

(ii) Given,

⇒ (5x - 4y)(5x + 3y)

⇒ 25x² + 15xy - 20xy - 12y²

⇒ 25x² - 5xy - 12y².

Hence, (5x - 4y)(5x + 3y) = 25x² - 5xy - 12y².

(iii) Given,

⇒ (3 - 7a)(3 + 4a)

⇒ 9 + 12a - 21a - 28a²

⇒ 9 - 9a - 28a².

Hence, (3 - 7a)(3 + 4a) = 9 - 9a - 28a².

We know that,

(a + b)(a - b) = a² - b²

(i) Given,

⇒ (3a + 2b)(3a - 2b)

⇒ (3a)² - (2b)²

⇒ 9a² - 4b².

Hence, (3a + 2b)(3a - 2b) = 9a² - 4b².

(ii) Given,

$$\begin{aligned}\Rightarrow \Big(5x + \frac{1}{5x}\Big)\Big(5x - \frac{1}{5x}\Big)\\[1em] \Rightarrow (5x)^2 - \Big(\frac{1}{5x}\Big)^2 \\[1em] \Rightarrow 25x^2 - \frac{1}{25x^2}.\end{aligned}$$

Hence, $\Big(5x + \frac{1}{5x}\Big)\Big(5x - \frac{1}{5x}\Big) = 25x^2 - \frac{1}{25x^2}$.

(iii) Given,

$$\begin{aligned}\Rightarrow \Big(2x^2 + \frac{3}{x^2}\Big)\Big(2x^2 - \frac{3}{x^2}\Big) \\[1em] \Rightarrow (2x^2)^2 - \Big(\frac{3}{x^2}\Big)^2 \\[1em] \Rightarrow 4x^4 - \frac{9}{x^4} \\[1em]\end{aligned}$$

Hence, $\Big(2x^2 + \frac{3}{x^2}\Big)\Big(2x^2 - \frac{3}{x^2}\Big) = 4x^4 - \frac{9}{x^4}$.

We know that,

(a + b)(a - b) = a² - b²

(i) Given,

⇒ (2 - x)(2 + x)(4 + x²)

⇒ [(2)² - (x)²](4 + x²)

⇒ (4 - x²)(4 + x²)

⇒ (4)² - (x²)²

⇒ 16 - x⁴.

Hence, (2 - x)(2 + x)(4 + x²) = 16 - x⁴.

(ii) Given,

⇒ (x + y)(x - y)(x² + y²)

⇒ [(x)² - (y)²](x² + y²)

⇒ (x² - y²)(x² + y²)

⇒ (x²)² - (y²)²

⇒ (x⁴ - y⁴).

Hence, (x + y)(x - y)(x² + y²) = (x⁴ - y⁴).

(i) Given,

⇒ (x - 2)(x - 3)(x + 4)

⇒ (x² - 3x - 2x + 6)(x + 4)

⇒ (x² - 5x + 6)(x + 4)

⇒ x²(x + 4) - 5x(x + 4) + 6(x + 4)

⇒ (x³ + 4x² - 5x² - 20x + 6x + 24)

⇒ (x³ - x² - 14x + 24)

Hence, (x - 2)(x - 3)(x + 4) = x³ - x² - 14x + 24.

(ii) Given,

⇒ (x - 5)(2x - 1)(2x + 3)

⇒ (2x² - x - 10x + 5)(2x + 3)

⇒ (2x² - 11x + 5)(2x + 3)

⇒ 2x²(2x + 3) - 11x(2x + 3) + 5(2x + 3)

⇒ (4x³ + 6x² - 22x² - 33x + 10x + 15)

⇒ (4x³ - 16x² - 23x + 15)

Hence, (x - 5)(2x - 1)(2x + 3) = 4x³ - 16x² - 23x + 15.

(i) Given,

⇒ (a + b)² + (a - b)²

⇒ a² + b² + 2ab + a² + b² - 2ab

⇒ 2a² + 2b²

⇒ 2(a² + b²)

Hence, (a + b)² + (a - b)² = 2(a² + b²).

(ii) Given,

⇒ (a + b)² - (a - b)²

⇒ (a² + b² + 2ab) - (a² + b² - 2ab)

⇒ a² + b² + 2ab - a² - b² + 2ab

⇒ 4ab

Hence, (a + b)² - (a - b)² = 4ab.

(iii) Given,

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{x}\Big)^2 + \Big(x - \frac{1}{x}\Big)^2 \\[1em] \Rightarrow \Big[x^2 + \Big(\frac{1}{x^2}\Big) + 2 \times x \times \Big(\frac{1}{x}\Big) + x^2 + \Big(\frac{1}{x^2}\Big) - 2 \times x \times \Big(\frac{1}{x}\Big)\Big] \\[1em] \Rightarrow \Big(x^2 + \frac{1}{x^2} + 2\Big) + \Big(x^2 + \frac{1}{x^2} - 2\Big) \\[1em] \Rightarrow 2x^2 + \frac{2}{x^2} \\[1em] \Rightarrow 2\Big(x^2 + \frac{1}{x^2}\Big) \\[1em]\end{aligned}$$

Hence, $\Big(x + \frac{1}{x}\Big)^2 + \Big(x - \frac{1}{x}\Big)^2 = 2\Big(x^2 + \frac{1}{x^2}\Big)$.

(iv) Given,

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{x}\Big)^2 - \Big(x - \frac{1}{x}\Big)^2 \\[1em] \Rightarrow \Big[x^2 + \Big(\frac{1}{x^2}\Big) + 2 \times x \times \Big(\frac{1}{x}\Big)\Big] - \Big[x^2 + \Big(\frac{1}{x^2}\Big) - 2 \times x \times \Big(\frac{1}{x}\Big)\Big] \\[1em] \Rightarrow \Big(x^2 + \frac{1}{x^2} + 2\Big) - \Big(x^2 + \frac{1}{x^2} - 2\Big) \\[1em] \Rightarrow x^2 + \frac{1}{x^2} + 2 - x^2 - \frac{1}{x^2} + 2 \\[1em] \Rightarrow 4.\end{aligned}$$

Hence, $\Big(x + \frac{1}{x}\Big)^2 - \Big(x - \frac{1}{x}\Big)^2 = 4$.

(v) Given,

$$\begin{aligned}\Rightarrow \Big(\frac{a}{2b} + \frac{2b}{a}\Big)^2 - \Big(\frac{2b}{a} - \frac{a}{2b}\Big)^2 \\[1em] \Rightarrow \Big[\Big(\frac{a}{2b}\Big)^2 + \Big(\frac{2b}{a}\Big)^2 + 2 \times \frac{a}{2b} \times \frac{2b}{a}\Big] - \Big[\Big(\frac{2b}{a}\Big)^2 + \Big(\frac{a}{2b}\Big)^2 - 2 \times \frac{a}{2b} \times \frac{2b}{a}\Big] \\[1em] \Rightarrow \Big(\frac{a^2}{4b^2} + \frac{4b^2}{a^2} + 2\Big) - \Big(\frac{4b^2}{a^2} + \frac{a^2}{4b^2} - 2\Big) \\[1em] \Rightarrow \frac{a^2}{4b^2} + \frac{4b^2}{a^2} + 2 - \frac{a^2}{4b^2} - \frac{4b^2}{a} + 2 \\[1em] \Rightarrow 4.\end{aligned}$$

Hence, $\Big(\frac{a}{2b} + \frac{2b}{a}\Big)^2 - \Big(\frac{2b}{a} - \frac{a}{2b}\Big)^2 = 4$.

(vi) Given,

$$\begin{aligned}\Rightarrow \Big(3x - \frac{1}{3x}\Big)^2 - \Big(3x + \frac{1}{3x}\Big)\Big(3x - \frac{1}{3x}\Big) \\[1em] \Rightarrow \Big[(3x)^2 + \Big(\frac{1}{3x}\Big)^2 - 2 \times 3x \times \frac{1}{3x}\Big] - \Big[(3x)^2 - \Big(\frac{1}{3x}\Big)^2\Big] \\[1em] \Rightarrow \Big(9x^2 + \frac{1}{9x^2} - 2 \Big) - \Big(9x^2 - \frac{1}{9x^2}\Big) \\[1em] \Rightarrow 9x^2 + \frac{1}{9x^2} - 2 - 9x^2 + \frac{1}{9x^2} \\[1em] \Rightarrow \frac{2}{9x^2} - 2 \\[1em] \Rightarrow 2\Big(\frac{1}{9x^2} - 1\Big)\end{aligned}$$

Hence, $\Big(3x - \frac{1}{3x}\Big)^2 - \Big(3x + \frac{1}{3x})\Big(3x - \frac{1}{3x}\Big) = 2\Big(\frac{1}{9x^2} - 1\Big)$.

(vii) Given,

⇒ (5a + 3b)² - (5a - 3b)² - 60ab

⇒ [(5a)² + (3b)² + 2 × 5a × 3b] - [(5a)² + (3b)² - 2 × 5a × 3b] - 60ab

⇒ [25a² + 9b² + 2 × 5a × 3b] - [25a² + 9b² - 2 × 5a × 3b] - 60ab

⇒ 25a² + 9b² + 30ab - 25a² - 9b² + 30ab - 60ab

⇒ 60ab - 60ab

⇒ 0

Hence, (5a + 3b)² - (5a - 3b)² - 60ab = 0.

(viii) Given,

⇒ (3x + 1)² - [(3x + 2)(3x - 1)]

⇒ (3x)² + (1)² + 2 × 3x × 1 - (9x² - 3x + 6x - 2)

⇒ 9x² + 1 + 6x - (9x² + 3x - 2)

⇒ 9x² + 1 + 6x - 9x² - 3x + 2

⇒ 1 + 3x + 2

⇒ 3x + 3

⇒ 3(x + 1).

Hence, (3x + 1)² - (3x + 2)(3x - 1) = 3(x + 1).

(i) Given,

(a + b) = 7 and ab = 10

Using identity,

⇒ (a + b)² - (a - b)² = 4ab

Substituting values we get :

⇒ (7)² - (a - b)² = 4 × 10

⇒ 49 - (a - b)² = 40

⇒ (a - b)² = 49 - 40

⇒ (a - b)² = 9

⇒ (a - b)² = $\sqrt{9}$

⇒ (a - b) = $\pm 3$

Hence, (a - b) = $\pm 3$.

(ii) Given,

(x - y) = 5 and xy = 24

Using identity,

⇒ (x + y)² - (x - y)² = 4xy

⇒ (x + y)² = 4xy + (x - y)²

⇒ (x + y)² = 4 × 24 + (5)²

⇒ (x + y)² = 96 + 25

⇒ (x + y) = $\sqrt{121}$

⇒ (x + y) = $\pm 11$

Hence, (x + y) = $\pm 11$.

⇒ (3a + 4b)² = (3a)² + (4b)² + 2 × 3a × 4b

⇒ (3a + 4b)² = 9a² + 16b² + 24ab

⇒ 9a² + 16b² = (3a + 4b)² - 24ab

Given,

(3a + 4b) = 16 and ab = 4

Substituting values we get :

⇒ 9a² + 16b² = (16)² - 24 × 4

⇒ 9a² + 16b² = 256 - 96

⇒ 9a² + 16b² = 160.

Hence, 9a² + 16b² = 160.

(i) Given,

(a + b) = 2 and (a - b) = 10

Using identity,

⇒ (a + b)² + (a - b)² = 2(a² + b²)

⇒ (2)² + (10)² = 2(a² + b²)

⇒ 2(a² + b²) = 4 + 100

⇒ 2(a² + b²) = 104

⇒ a² + b² = 52.

Hence, a² + b² = 52.

(ii) Given,

(a + b) = 2 and (a - b) = 10

Using identity,

⇒ (a + b)² - (a - b)² = 4ab

⇒ (2)² - (10)² = 4ab

⇒ 4ab = 4 - 100

⇒ 4ab = -96

⇒ ab = -24

Hence, ab = -24.

(i) Given,

(a - b) = 0.9 and ab = 0.36

Using identity,

⇒ (a + b)² - (a - b)² = 4ab

⇒ (a + b)² = 4ab + (a - b)²

⇒ (a + b)² = 4 × 0.36 + (0.9)²

⇒ (a + b)² = 1.44 + 0.81

⇒ (a + b)² = 2.25

⇒ (a + b) = $\sqrt{2.25}$

⇒ (a + b) = $\pm 1.5$

Hence, (a + b) = $\pm 1.5$

(ii) Using identity,

⇒ a² − b² = (a − b)(a + b)

⇒ a² − b² = 0.9 × $\pm 1.5$

⇒ a² − b² = $\pm 1.35$

Hence, a² − b² = $\pm 1.35$

(i) Given,

$\Big(x + \frac{1}{x}\Big) = 5$

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{x}\Big)^2 = x^2 + \Big(\frac{1}{x}\Big)^2 + 2 \times x \times \frac{1}{x} \\[1em] \Rightarrow (5)^2 = x^2 + \frac{1}{x^2} + 2 \times x \times \frac{1}{x} \\[1em] \Rightarrow 25 = x^2 + \frac{1}{x^2} + 2 \\[1em] \Rightarrow x^2 + \frac{1}{x^2} = 25 - 2 \\[1em] \Rightarrow x^2 + \frac{1}{x^2} = 23.\end{aligned}$$

Hence, $x^2 + \frac{1}{x^2} = 23$.

(ii) Given,

$\Big(x + \frac{1}{x}\Big) = 5$

From part (i),

$x^2 + \frac{1}{x^2} = 23$

Using identity,

$$\begin{aligned}\Rightarrow \Big(x^2 + \frac{1}{x^2}\Big)^2 = (x^2)^2 + \Big(\frac{1}{x^2}\Big)^2 + 2 \times x^2 \times \frac{1}{x^2} \\[1em] \Rightarrow (23)^2 = (x^2)^2 + \Big(\frac{1}{x^2}\Big)^2 + 2 \times x^2 \times \frac{1}{x^2} \\[1em] \Rightarrow 529 = x^4 + \frac{1}{x^4} + 2 \\[1em] \Rightarrow x^4 + \frac{1}{x^4} = 529 - 2 \\[1em] \Rightarrow x^4 + \frac{1}{x^4} = 527 \\[1em]\end{aligned}$$

Hence, $x^4 + \frac{1}{x^4} = 527$.

(i) Given,

$\Big(x - \frac{1}{x}\Big) = 4$

$$\begin{aligned}\Rightarrow \Big(x - \frac{1}{x}\Big)^2 = x^2 + \Big(\frac{1}{x}\Big)^2 - 2 \times x \times \frac{1}{x} \\[1em] \Rightarrow (4)^2 = x^2 + \Big(\frac{1}{x}\Big)^2 - 2 \times x \times \frac{1}{x} \\[1em] \Rightarrow 16 = x^2 + \frac{1}{x^2} - 2 \\[1em] \Rightarrow x^2 + \frac{1}{x^2} = 16 + 2 \\[1em] \Rightarrow x^2 + \frac{1}{x^2} = 18\end{aligned}$$

Hence, $x^2 + \frac{1}{x^2} = 18$.

(ii) Given,

$\Big(x - \frac{1}{x}\Big) = 4$

From part (i),

$x^2 + \frac{1}{x^2} = 18$

$$\begin{aligned}\Rightarrow \Big(x^2 + \frac{1}{x^2}\Big)^2 = (x^2)^2 + \Big(\frac{1}{x^2}\Big)^2 + 2 \times x^2 \times \frac{1}{x^2} \\[1em] \Rightarrow (18)^2 = (x^2)^2 + \Big(\frac{1}{x^2}\Big)^2 + 2 \times x^2 \times \frac{1}{x^2} \\[1em] \Rightarrow 324 = x^4 + \frac{1}{x^4} + 2 \\[1em] \Rightarrow x^4 + \frac{1}{x^4} = 324 - 2 \\[1em] \Rightarrow x^4 + \frac{1}{x^4} = 322.\end{aligned}$$

Hence, $x^4 + \frac{1}{x^4} = 322$.

(i) Given,

$$\begin{aligned}\Rightarrow x - 2 = \frac{1}{3x} \\[1em] \Rightarrow x - \frac{1}{3x} = 2\end{aligned}$$

We know that,

$$\begin{aligned}\Rightarrow \Big(x - \frac{1}{3x}\Big)^2 = x^2 + \Big(\frac{1}{3x}\Big)^2 - 2 \times x \times \frac{1}{3x} \\[1em] \Rightarrow (2)^2 = x^2 + \Big(\frac{1}{3x}\Big)^2 - 2 \times x \times \frac{1}{3x} \\[1em] \Rightarrow 4 = x^2 + \frac{1}{9x^2} - \frac{2}{3} \\[1em] \Rightarrow x^2 + \frac{1}{9x^2} = 4 + \frac{2}{3} \\[1em] \Rightarrow x^2 + \frac{1}{9x^2} = \frac{12 + 2}{3} \\[1em] \Rightarrow x^2 + \frac{1}{9x^2} = \frac{14}{3}\end{aligned}$$

Hence, $x^2 + \frac{1}{9x^2} = \frac{14}{3}$.

(ii) From part (i),

$x^2 + \frac{1}{9x^2} = \frac{14}{3}$

Using identity,

$$\begin{aligned}\Rightarrow \Big(x^2 + \frac{1}{9x^2}\Big)^2 = (x^2)^2 + \Big(\frac{1}{9x^2}\Big)^2 + 2 \times x^2 \times \frac{1}{9x^2} \\[1em] \Rightarrow \Big(\frac{14}{3}\Big)^2 = (x^2)^2 + \Big(\frac{1}{9x^2}\Big)^2 + 2 \times x^2 \times \frac{1}{9x^2} \\[1em] \Rightarrow \frac{196}{9} = x^4 + \frac{1}{81x^4} + \frac{2}{9} \\[1em] \Rightarrow x^4 + \frac{1}{81x^4} = \frac{196}{9} - \frac{2}{9} \\[1em] \Rightarrow x^4 + \frac{1}{81x^4} = \frac{196 - 2}{9} \\[1em] \Rightarrow x^4 + \frac{1}{81x^4} = \frac{194}{9}\end{aligned}$$

Hence, $x^4 + \frac{1}{81x^4} = \frac{194}{9}$.

(i) Given,

$\Big(x + \frac{1}{x}\Big) = 6$

We know that,

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{x}\Big)^2 - \Big(x - \frac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow (6)^2 - \Big(x - \frac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow 36 - 4 = \Big(x - \frac{1}{x}\Big)^2 \\[1em] \Rightarrow 32 = \Big(x - \frac{1}{x}\Big)^2 \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big) = \sqrt{32} \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big) = \pm 4\sqrt{2}.\end{aligned}$$

Hence, $\Big(x - \frac{1}{x}\Big) = \pm 4\sqrt{2}$.

(ii) Given,

$\Big(x + \frac{1}{x}\Big) = 6$

From part (i),

$\Rightarrow \Big(x - \frac{1}{x}\Big) = ±4\sqrt{2}$

We know that,

$$\begin{aligned}\Rightarrow \Big(x^2 - \frac{1}{x^2}\Big) = \Big(x + \frac{1}{x}\Big)\Big(x - \frac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^2 - \frac{1}{x^2}\Big) = 6 \times \pm 4\sqrt{2} \\[1em] \Rightarrow \Big(x^2 - \frac{1}{x^2}\Big) = \pm 24\sqrt{2}\end{aligned}$$

Hence, $x^2 - \frac{1}{x^2} = \pm 24\sqrt{2}$.

(i) Given,

$\Big(x - \frac{1}{x}\Big) = 8$

We know that,

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{x}\Big)^2 - \Big(x - \frac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(x + \frac{1}{x}\Big)^2 - (8)^2 = 4 \\[1em] \Rightarrow \Big(x + \frac{1}{x}\Big)^2 - 64 = 4 \\[1em] \Rightarrow \Big(x + \frac{1}{x}\Big)^2 = 64 + 4 \\[1em] \Rightarrow \Big(x + \frac{1}{x}\Big)^2 = 68 \\[1em] \Rightarrow \Big(x + \frac{1}{x}\Big) = \sqrt{68} \\[1em] \Rightarrow \Big(x + \frac{1}{x}\Big) = \pm 2\sqrt{17} \\[1em]\end{aligned}$$

Hence, $\Big(x + \frac{1}{x}\Big) = \pm 2\sqrt{17}$.

(ii) Given,

$\Big(x - \frac{1}{x}\Big) = 8$

From (i),

$\Big(x + \frac{1}{x}\Big) = \pm 2\sqrt{17}$

Using identity,

$$\begin{aligned}\Rightarrow \Big(x^2 - \frac{1}{x^2}\Big) = \Big(x + \frac{1}{x}\Big)\Big(x - \frac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^2 - \frac{1}{x^2}\Big) = (\pm 2\sqrt{17})\times 8 \\[1em] \Rightarrow \Big(x^2 - \frac{1}{x^2}\Big) = \pm 16\sqrt{17}\end{aligned}$$

Hence, $x^2 - \frac{1}{x^2} = \pm 16\sqrt{17}$.

(i) Given,

$\Big(x^2 + \frac{1}{x^2}\Big) = 7$

Using identity,

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{x}\Big)^2 = x^2 + \frac{1}{x^2} + 2 \\[1em] \Rightarrow \Big(x + \frac{1}{x}\Big)^2 = 7 + 2 \\[1em] \Rightarrow \Big(x + \frac{1}{x}\Big)^2 = 9 \\[1em] \Rightarrow \Big(x + \frac{1}{x}\Big) = \pm \sqrt{9} \\[1em] \Rightarrow \Big(x + \frac{1}{x}\Big) = \pm 3.\end{aligned}$$

Hence, $\Big(x + \frac{1}{x}\Big) = \pm 3$.

(ii) Given,

$\Big(x^2 + \frac{1}{x^2}\Big) = 7$

Using identity,

$$\begin{aligned}\Rightarrow \Big(x - \frac{1}{x}\Big)^2 = x^2 + \frac{1}{x^2} - 2 \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big)^2 = 7 - 2 \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big)^2 = 5 \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big) = \pm \sqrt{5} \\[1em]\end{aligned}$$

Hence, $\Big(x - \frac{1}{x}\Big) = \pm \sqrt{5}$.

(iii) Given,

$\Big(x^2 + \frac{1}{x^2}\Big) = 7$

From part (i) and (ii),

$\Big(x + \frac{1}{x}\Big) = \pm 3 \text{ and } \Big(x - \frac{1}{x}\Big) = \pm \sqrt{5}$

Using identity,

$$\begin{aligned}\Rightarrow \Big(x^2 - \frac{1}{x^2}\Big) = \Big(x + \frac{1}{x}\Big)\Big(x - \frac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^2 - \frac{1}{x^2}\Big) = (\pm 3) \times (\pm \sqrt{5}) \\[1em] \Rightarrow \Big(x^2 - \frac{1}{x^2}\Big) = \pm 3\sqrt{5} \\[1em] \Rightarrow 2\Big(x^2 - \frac{1}{x^2}\Big) = 2 \times \pm 3\sqrt{5} \\[1em] \Rightarrow \Big(2x^2 - \frac{2}{x^2}\Big) = 2 \times (\pm 3\sqrt{5}) \\[1em] \Rightarrow \Big(2x^2 - \frac{2}{x^2}\Big) = \pm 6\sqrt{5}\end{aligned}$$

Hence, $\Big(2x^2 - \frac{2}{x^2}\Big) = \pm 6\sqrt{5}$.

$$\begin{aligned}\Rightarrow \Big(x - \frac{1}{5x}\Big)^2 = \Big[x^2 + \Big(\frac{1}{5x}\Big)^2 - 2 \times x \times \frac{1}{5x}\Big] \\[1em] \Rightarrow \Big(x - \frac{1}{5x}\Big)^2 = \Big[x^2 + \frac{1}{25x^2} - \frac{2}{5}\Big] \\[1em] \Rightarrow \Big(x - \frac{1}{5x}\Big)^2 = 9\frac{2}{5} - \frac{2}{5} \\[1em] \Rightarrow \Big(x - \frac{1}{5x}\Big)^2 = \frac{47}{5} - \frac{2}{5} \\[1em] \Rightarrow \Big(x - \frac{1}{5x}\Big)^2 = \frac{47 - 2}{5} \\[1em] \Rightarrow \Big(x - \frac{1}{5x}\Big)^2 = \frac{45}{5} \\[1em] \Rightarrow \Big(x - \frac{1}{5x}\Big)^2 = 9 \\[1em] \Rightarrow (x - \frac{1}{5x}\Big) = \sqrt{9} \\[1em] \Rightarrow \Big(x - \frac{1}{5x}\Big) = \pm 3 \\[1em]\end{aligned}$$

Hence, $\Big(x - \frac{1}{5x}\Big) = \pm 3$.

(i) Solving,

$$\begin{aligned}\Rightarrow a^2 - 4a - 1 = 0 \\[1em] \Rightarrow a^2 - 4a = 1 \\[1em] \Rightarrow a(a - 4) = 1 \\[1em] \Rightarrow a - 4 = \frac{1}{a} \\[1em] \Rightarrow a - \frac{1}{a} = 4 \\[1em]\end{aligned}$$

Hence, $\Big(a - \frac{1}{a}\Big) = 4$

(ii) Using identity,

$$\begin{aligned}\Rightarrow \Big(a + \frac{1}{a}\Big)^2 - \Big(a - \frac{1}{a}\Big)^2 = 4 \\[1em] \Rightarrow \Big(a + \frac{1}{a}\Big)^2 - (4)^2 = 4 \\[1em] \Rightarrow \Big(a + \frac{1}{a}\Big)^2 - 16 = 4 \\[1em] \Rightarrow \Big(a + \frac{1}{a}\Big)^2 = 4 + 16\\[1em] \Rightarrow \Big(a + \frac{1}{a}\Big)^2 = 20 \\[1em] \Rightarrow a + \frac{1}{a} = \sqrt{20} \\[1em] \Rightarrow a + \frac{1}{a} = \pm 2\sqrt{5}.\end{aligned}$$

Hence, $a + \frac{1}{a} = \pm 2\sqrt{5}$

(iii) From part (i) and (ii),

$$\begin{aligned}\Rightarrow a - \frac{1}{a} = 4 \\[1em] \Rightarrow a + \frac{1}{a} = \pm 2\sqrt{5}\end{aligned}$$

Case 1:

$\Rightarrow a + \frac{1}{a} = 2\sqrt{5}$

Using identity,

$$\begin{aligned}\Rightarrow \Big(a + \frac{1}{a}\Big)\Big(a - \frac{1}{a}\Big) = \Big( a^2 - \frac{1}{a^2}\Big) \\[1em] \Rightarrow (2\sqrt{5}) \times (4) = \Big( a^2 - \frac{1}{a^2}\Big) \\[1em] \Rightarrow \Big( a^2 - \frac{1}{a^2}\Big) = 8\sqrt{5} \\[1em]\end{aligned}$$

Case 2:

$\Rightarrow a + \frac{1}{a} = -2\sqrt{5}$

Using identity,

$$\begin{aligned}\Rightarrow \Big(a + \frac{1}{a}\Big)\Big(a - \frac{1}{a}\Big) = \Big( a^2 - \frac{1}{a^2}\Big) \\[1em] \Rightarrow (-2\sqrt{5}) \times (4) = \Big( a^2 - \frac{1}{a^2}\Big) \\[1em] \Rightarrow \Big( a^2 - \frac{1}{a^2}\Big) = -8\sqrt{5} \\[1em]\end{aligned}$$

Hence, $\Big( a^2 - \frac{1}{a^2}\Big) = \pm 8\sqrt{5}$.

(iv) From part (i) and (ii),

$$\begin{aligned}\Rightarrow a - \frac{1}{a} = 4 \\[1em] \Rightarrow a + \frac{1}{a} = \pm 2\sqrt{5}\end{aligned}$$

Using identity,

$$\begin{aligned}\Rightarrow \Big(a + \frac{1}{a}\Big)^2 + \Big(a - \frac{1}{a}\Big)^2 = 2\Big( a^2 + \frac{1}{a^2}\Big) \\[1em] \Rightarrow (\pm 2\sqrt{5})^2 + (4)^2 = 2\Big( a^2 + \frac{1}{a^2}\Big) \\[1em] \Rightarrow 2\Big( a^2 + \frac{1}{a^2}\Big) = 20 + 16 \\[1em] \Rightarrow 2\Big( a^2 + \frac{1}{a^2}\Big) = 36 \\[1em] \Rightarrow \Big( a^2 + \frac{1}{a^2}\Big) = \frac{36}{2} \\[1em] \Rightarrow \Big( a^2 + \frac{1}{a^2}\Big) = 18 \\[1em]\end{aligned}$$

Hence, $\Big(a^2 + \frac{1}{a^2}\Big) = 18$.

(i) Given,

$a = \frac{1}{a - 5}$

$$\begin{aligned}\Rightarrow a = \frac{1}{a - 5} \\[1em] \Rightarrow a - 5 = \frac{1}{a} \\[1em] \Rightarrow a - \frac{1}{a} = 5\end{aligned}$$

Hence, $\Big(a - \frac{1}{a}\Big) = 5$

(ii) From part (i),

$a - \frac{1}{a} = 5$

Using identity,

$$\begin{aligned}\Rightarrow \Big(a + \frac{1}{a}\Big)^2 - \Big(a - \frac{1}{a}\Big)^2 = 4 \\[1em] \Rightarrow \Big(a + \frac{1}{a}\Big)^2 - (5)^2 = 4 \\[1em] \Rightarrow \Big(a + \frac{1}{a}\Big)^2 = 4 + 25\\[1em] \Rightarrow \Big(a + \frac{1}{a}\Big)^2 = 29 \\[1em] \Rightarrow a + \frac{1}{a} = \pm \sqrt{29}\end{aligned}$$

Hence, $a + \frac{1}{a} = \pm \sqrt{29}$

(iii) From (i) and (ii),

$$\begin{aligned}\Rightarrow a - \frac{1}{a} = 5 \\[1em] \Rightarrow a + \frac{1}{a} = \pm \sqrt{29}\end{aligned}$$

Using identity,

$\Big(a + \frac{1}{a}\Big)\Big(a - \frac{1}{a}\Big) = \Big( a^2 - \frac{1}{a^2}\Big)$

$$\begin{aligned}\Rightarrow (\pm \sqrt{29}) \times (5) = \Big( a^2 - \frac{1}{a^2}\Big) \\[1em] \Rightarrow \Big( a^2 - \frac{1}{a^2}\Big) = \pm 5\sqrt{29} \\[1em]\end{aligned}$$

Hence, $\Big( a^2 - \frac{1}{a^2}\Big) = \pm 5\sqrt{29}$

(iv) From (i) and (ii),

$$\begin{aligned}\Rightarrow a - \frac{1}{a} = 5 \\[1em] \Rightarrow a + \frac{1}{a} = \pm\sqrt{29}\end{aligned}$$

Using identity,

$\Big(a + \frac{1}{a}\Big)^2 + \Big(a - \frac{1}{a}\Big)^2 = 2\Big( a^2 + \frac{1}{a^2}\Big)$

$$\begin{aligned}\Rightarrow (\pm \sqrt{29})^2 + (5)^2 = 2\Big( a^2 + \frac{1}{a^2}\Big) \\[1em] \Rightarrow 2\Big( a^2 + \frac{1}{a^2}\Big) = 29 + 25 \\[1em] \Rightarrow 2\Big( a^2 + \frac{1}{a^2}\Big) = 54 \\[1em] \Rightarrow \Big( a^2 + \frac{1}{a^2}\Big) = \frac{54}{2} \\[1em] \Rightarrow \Big( a^2 + \frac{1}{a^2}\Big) = 27 \\[1em]\end{aligned}$$

Hence, $\Big(a^2 + \frac{1}{a^2}\Big) = 27$.

(i) Given,

⇒ (137)²

⇒ (130 + 7)²

Using identity :

(a + b)² = a² + b² + 2ab

⇒ (130 + 7)² = (130)² + 7² + 2 × 130 × 7

⇒ (130 + 7)² = 16900 + 49 + 1820 = 18769.

Hence, (137)² = 18769.

(ii) Given,

⇒ (1008)²

⇒ (1000 + 8)²

Using identity :

(a + b)² = a² + b² + 2ab

⇒ (1000 + 8)² = (1000)² + 8² + 2 × 1000 × 8

⇒ (1000 + 8)² = 1000000 + 64 + 16000 = 1016064.

Hence, (1008)² = 1016064.

(iii) Given,

⇒ (11.6)²

⇒ (11 + 0.6)²

Using identity :

(a + b)² = a² + b² + 2ab

⇒ (11 + 0.6)² = (11)² + (0.6)² + 2 × 11 × 0.6

⇒ (11 + 0.6)² = 121 + 0.36 + 13.2 = 134.56

Hence, (11.6)² = 134.56.

(i) Given,

⇒ (97)²

⇒ (100 - 3)²

Using identity :

⇒ (a - b)² = a² + b² - 2ab

⇒ (100 - 3)² = (100)² + 3² - 2 × 100 × 3

⇒ (100 - 3)² = 10000 + 9 - 600

⇒ 9409.

Hence, (97)² = 9409.

(ii) Given,

⇒ (992)²

⇒ (1000 - 8)²

Using identity :

(a - b)² = a² + b² - 2ab

⇒ (1000 - 8)² = (1000)² + 8² - 2 × 1000 × 8

⇒ (1000 - 8)² = 1000000 + 64 - 16000

⇒ 984064.

Hence, (992)² = 984064.

(iii) Given,

⇒ (9.98)²

⇒ (10 - 0.02)²

Using identity :

(a - b)² = a² + b² - 2ab

⇒ (10 - 0.02)² = (10)² + 0.02² - 2 × 10 × 0.02

⇒ (10 - 0.02)² = 100 + 0.0004 - 0.4

⇒ 99.6004

Hence, (9.98)² = 99.6004.

(i) Given,

16a² + 9b² + ..............

Adding 24ab to above equation, we get :

⇒ 16a² + 9b² + 24ab

⇒ (4a)² + (3b)² + 2 × 4a × 3b

We know that,

(a + b)² = a² + b² + 2ab

⇒ (4a + 3b)².

Hence, on adding 24ab to the expression 16a² + 9b², it becomes a perfect square.

(ii) Given,

25a² + 16b² + ..............

Adding 40ab to above equation, we get :

⇒ 25a² + 16b² + 40ab

⇒ (5a)² + (4b)² + 2 × 5a × 4b

We know that,

(a + b)² = a² + b² + 2ab

⇒ (5a + 4b)².

Hence, on adding 40ab to the expression 25a² + 16b², it becomes a perfect square.

(iii) Given,

4a² + 20ab + ..............

Adding 25b² to above equation, we get :

⇒ 4a² + 25b² + 20ab

⇒ (2a)² + (5b)² + 2 × 2a × 5b

We know that,

(a + b)² = a² + b² + 2ab

⇒ (2a + 5b)²

Hence, on adding 25b² to the expression 4a² + 20ab, it becomes a perfect square.

(iv) Given,

9a² - 24ab + ..............

Adding 16b² to above equation, we get :

⇒ 9a² + 16b² - 24ab

⇒ (3a)² + (4b)² - 2 × 3a × 4b

We know that,

(a - b)² = a² + b² - 2ab

⇒ (3a - 4b)²

Hence, on adding 16b² to the expression 9a² - 24ab, it becomes a perfect square.

Given,

(a + b + c) = 14

(a² + b² + c²) = 74

Using identity,

⇒ (a + b + c)² = (a² + b² + c²) + 2 (ab + bc + ca)

⇒ (14)² = (74) + 2 (ab + bc + ca)

⇒ 196 - 74 = 2 (ab + bc + ca)

⇒ 2 (ab + bc + ca) = 122

⇒ (ab + bc + ca) = $\frac{122}{2}$

⇒ (ab + bc + ca) = 61.

Hence, (ab + bc + ca) = 61.

Given,

(a + b + c) = 15

(ab + bc + ca) = 74

Using identity,

⇒ (a + b + c)² = (a² + b² + c²) + 2 (ab + bc + ca)

⇒ (15)² = (a² + b² + c²) + 2 (74)

⇒ 225 = (a² + b² + c²) + 148

⇒ 225 - 148 = (a² + b² + c²)

⇒ (a² + b² + c²) = 77

Hence, (a² + b² + c²) = 77.

Given,

(a² + b² + c²) = 50

(ab + bc + ca) = 47

Using identity,

⇒ (a + b + c)² = (a² + b² + c²) + 2 (ab + bc + ca)

⇒ (a + b + c)² = (50) + 2 (47)

⇒ (a + b + c)² = 50 + 94

⇒ (a + b + c)² = 144

⇒ (a + b + c) = $\sqrt{144}$

⇒ (a + b + c) = $\pm 12$

Hence, (a + b + c) = $\pm 12$.

Given,

(a² + b² + c²) = 89

(ab - bc - ca) = 16

Using identity,

⇒ (a + b - c)² = (a² + b² + c²) + 2 (ab - bc - ca)

⇒ (a + b - c)² = (89) + 2 (16)

⇒ (a + b - c)² = 89 + 32

⇒ (a + b - c)² = 121

⇒ (a + b - c) = $\sqrt{121}$

⇒ (a + b - c) = $\pm 11$

Hence, (a + b - c) = $\pm 11$.

(i) Given,

(3a + 5b)³

Using identity :

(a + b)³ = a³ + b³ + 3ab(a + b)

⇒ (3a + 5b)³ = (3a)³ + (5b)³ + 3 × 3a × 5b × (3a + 5b)

⇒ (3a + 5b)³ = 27a³ + 125b³ + 45ab × (3a + 5b)

⇒ (3a + 5b)³ = 27a³ + 135a² b + 225ab² + 125b³

Hence, (3a + 5b)³ = 27a³ + 135a²b + 225ab² + 125b³.

(ii) Given,

(2p - 3q)³

Using identity :

(a - b)³ = a³ - b³ - 3a²b + 3ab²

⇒ (2p - 3q)³ = [(2p)³ - (3q)³ - 3 × (2p)² (3q) + 3 × (2p) × (3q)²]

⇒ (2p - 3q)³ = 8p³ - 27q³ - 3 × 4p² (3q) + 3 × (2p) × 9q²

⇒ (2p - 3q)³ = 8p³ - 36p²q + 54pq² - 27q³

Hence, (2p - 3q)³ = 8p³ - 36p²q + 54pq² - 27q³.

(iii) Given,

$\Big(2x + \frac{1}{3x}\Big)^3$

Using identity :

(a + b)³ = a³ + b³ + 3a²b + 3ab²

$$\begin{aligned}\Rightarrow \Big(2x + \frac{1}{3x}\Big)^3 = \Big[(2x)^3 + \Big(\frac{1}{3x}\Big)^3 + 3 \times (2x)^2 \times \Big(\frac{1}{3x}\Big) + 3 \times 2x \times \Big(\frac{1}{3x}\Big)^2 \Big]\\[1em] \Rightarrow \Big(2x + \frac{1}{3x}\Big)^3 = 8x^3 + \frac{1}{27x^3} + 3 \times 4x^2 \times \Big(\frac{1}{3x}\Big) + 3 \times 2x \times \Big(\frac{1}{9x^2}\Big) \\[1em] \Rightarrow \Big(2x + \frac{1}{3x}\Big)^3 = 8x^3 + 4x + \frac{2}{3x} + \frac{1}{27x^3} \\[1em]\end{aligned}$$

Hence, $\Big(2x + \frac{1}{3x}\Big)^3 = 8x^3 + 4x + \frac{2}{3x} + \frac{1}{27x^3}$.

(iv) Given,

(3ab - 2c)³

Using identity :

(a - b)³ = a³ - b³ - 3a²b + 3ab²

⇒ (3ab - 2c)³ = [(3ab)³ - (2c)³ - 3 × (3ab)² (2c) + 3 × (3ab) × (2c)²]

⇒ (3ab - 2c)³ = 27a³b³ - 8c³ - (9a²b²) × 6c + 3 × (3ab) × 4c²

⇒ (3ab - 2c)³ = 27a³b³ - 54a²b²c + 36 abc² - 8c³

Hence, (3ab - 2c)³ = 27a³b³ - 54a²b²c + 36 abc² - 8c³.

(v) Given,

$\Big(3a - \frac{1}{a}\Big)^3$

Using identity :

(a - b)³ = a³ - b³ - 3a²b + 3ab²

$$\begin{aligned}\Rightarrow \Big(3a - \frac{1}{a}\Big)^3 = \Big[(3a)^3 - \Big(\frac{1}{a}\Big)^3 - 3 \times (3a)^2 \times \Big(\frac{1}{a}\Big) + 3 \times 3a \times \Big(\frac{1}{a}\Big)^2 \Big]\\[1em] \Rightarrow \Big(3a - \frac{1}{a}\Big)^3 = 27a^3 - \frac{1}{a^3} - 3 \times 9a^2 \times \Big(\frac{1}{a}\Big) + 3 \times 3a \times \Big(\frac{1}{a^2}\Big) \\[1em] \Rightarrow \Big(3a - \frac{1}{a}\Big)^3 = 27a^3 - 27a + \frac{9}{a} - \frac{1}{a^3} \\[1em]\end{aligned}$$

Hence, $\Big(3a - \frac{1}{a}\Big)^3 = 27a^3 - 27a + \frac{9}{a} - \frac{1}{a^3}$.

(vi) Given,

$\Big(\frac{1}{2}x - \frac{2}{3}y\Big)^3$

Using identity :

(a - b)³ = a³ - b³ - 3a²b + 3ab²

$$\begin{aligned}\Rightarrow \Big(\frac{1}{2}x - \frac{2}{3}y\Big)^3 = \Big[\Big(\frac{1}{2}x\Big)^3 - \Big(\frac{2}{3}y\Big)^3 - 3 \times \Big(\frac{1}{2}x\Big)^2 \times \Big(\frac{2}{3}y\Big) + 3 \times \Big(\frac{1}{2}x\Big) \times \Big(\frac{2}{3}y\Big)^2 \Big]\\[1em] \Rightarrow \Big(\frac{1}{2}x - \frac{2}{3}y\Big)^3 = \Big(\frac{1}{8}x^3\Big) - \Big(\frac{8}{27}y^3\Big) - 3 \times \Big(\frac{1}{4}x^2\Big) \times \Big(\frac{2}{3}y\Big) + 3 \times \Big(\frac{1}{2}x\Big) \times \Big(\frac{4}{9}y^2\Big) \\[1em] \Rightarrow \Big(\frac{1}{2}x - \frac{2}{3}y\Big)^3 = \frac{1}{8}x^3 - \frac{8}{27}y^3 - \frac{6}{12}x^2y + \frac{12}{18}xy^2 \\[1em] \Rightarrow \Big(\frac{1}{2}x - \frac{2}{3}y\Big)^3 = \frac{1}{8}x^3 - \frac{8}{27}y^3 - \frac{1}{2}x^2y + \frac{2}{3}xy^2 \\[1em]\end{aligned}$$

Hence, $\Big(\frac{1}{2}x - \frac{2}{3}y\Big)^3 = \frac{1}{8}x^3 - \frac{8}{27}y^3 - \frac{1}{2}x^2y + \frac{2}{3}xy^2$.

Given,

(4a + 3b) = 10

ab = 2

⇒ (4a + 3b)³ = [(4a)³ + (3b)³ + 3 × 4a × 3b × (4a + 3b)]

⇒ (10)³ = 64a³ + 27b³ + 36ab × (10)

⇒ 1000 = 64a³ + 27b³ + 36 × 2 × 10

⇒ 1000 = 64a³ + 27b³ + 720

⇒ 64a³ + 27b³ = 1000 - 720

⇒ 64a³ + 27b³ = 280

Hence, 64a³ + 27b³ = 280.

Given,

3x – 2y = 5

xy = 6

Using identity :

(a - b)³ = a³ - b³ - 3ab(a - b)

⇒ (3x – 2y)³ = [(3x)³ - (2y)³ - 3 × 3x × 2y × (3x - 2y)]

⇒ (5)³ = 27x³ - 8y³ - 18xy × (5)

⇒ 125 = 27x³ - 8y³ - 90xy

⇒ 125 = 27x³ - 8y³ - 90(6)

⇒ 27x³ - 8y³ = 125 + 540

⇒ 27x³ - 8y³ = 665

Hence, 27x³ - 8y³ = 665.

Given,

a + 3b = 6

⇒ (a + 3b)³ = [(a)³ + (3b)³ + 3 × a × 3b × (a + 3b)]

⇒ (6)³ = a³ + 27b³ + 9ab × (a + 3b)

⇒ 216 = a³ + 27b³ + 9ab × (6)

⇒ a³ + 27b³ + 54ab = 216

Hence proved that a³ + 27b³ + 54ab = 216.

We know that,

If x + y + z = 0 then x³ + y³ + z³ = 3xyz ........(1)

Since, (a + 2b + 3c) = 0,

⇒ (a)³ + (2b)³ + (3c)³ = 3 × a × 2b × 3c

⇒ a³ + 8b³ + 27c³ = 18abc.

Hence, proved that a³ + 8b³ + 27c³ = 18abc.

Given,

$x + \frac{1}{x} = 3$

Using identity,

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{x}\Big)^3 = x^3 + \frac{1}{x^3} + 3\Big(x + \frac{1}{x}\Big) \\[1em] \Rightarrow 3^3 = x^3 + \frac{1}{x^3} + 3 \times 3 \\[1em] \Rightarrow 27 = x^3 + \frac{1}{x^3} + 9 \\[1em] \Rightarrow x^3 + \frac{1}{x^3} = 27 - 9 = 18.\end{aligned}$$

Hence, $x^3 + \frac{1}{x^3} = 18.$

Given,

$x - \frac{1}{x} = 5$

Using identity,

$$\begin{aligned}\Rightarrow \Big(x^3 - \frac{1}{x^3}\Big) = \Big(x - \frac{1}{x}\Big)^3 + 3\Big(x - \frac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^3 - \frac{1}{x^3}\Big) = (5)^3 + 3 \times 5 \\[1em] \Rightarrow \Big(x^3 - \frac{1}{x^3}\Big) = 125 + 15 \\[1em] \Rightarrow \Big(x^3 - \frac{1}{x^3}\Big) = 140\end{aligned}$$

Hence, $x^3 - \frac{1}{x^3} = 140.$

Given,

$\Rightarrow x - \frac{2}{x} = 6$

Upon cubing both sides we get :

$$\begin{aligned}\Rightarrow \Big(x - \frac{2}{x}\Big)^3 = 6^3 \\[1em] \Rightarrow (x)^3 - \Big(\frac{2}{x}\Big)^3 - 3 \times x \times \frac{2}{x} \times \Big(x - \frac{2}{x}\Big) = 216 \\[1em] \Rightarrow (x)^3 - \Big(\frac{2}{x}\Big)^3 - 6 \times 6 = 216 \\[1em] \Rightarrow x^3 - \frac{8}{x^3} - 36 = 216 \\[1em] \Rightarrow x^3 - \frac{8}{x^3} = 216 + 36 \\[1em] \Rightarrow x^3 - \frac{8}{x^3} = 252.\end{aligned}$$

Hence, $x^3 - \frac{8}{x^3} = 252.$

(i) Given,

$x + \frac{1}{x} = 4$

Using identity,

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{x}\Big)^3 = x^3 + \frac{1}{x^3} + 3\Big(x + \frac{1}{x}\Big) \\[1em] \Rightarrow 4^3 = x^3 + \frac{1}{x^3} + 3 \times 4 \\[1em] \Rightarrow 64 = x^3 + \frac{1}{x^3} + 12 \\[1em] \Rightarrow x^3 + \frac{1}{x^3} = 64 - 12 = 52.\end{aligned}$$

Hence, $x^3 + \frac{1}{x^3} = 52.$

(ii) Given,

$x + \frac{1}{x} = 4$

Using identity,

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{x}\Big)^2 - \Big(x - \frac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow (4)^2 - \Big(x - \frac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow 16 - \Big(x - \frac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big)^2 = 16 - 4 \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big)^2 = 12 \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big) = \sqrt{12}\\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big) = \pm 2\sqrt{3}.\end{aligned}$$

Hence, $\Big(x - \frac{1}{x}\Big) = \pm 2\sqrt{3}.$

(iii) Given,

$\Big(x - \frac{1}{x}\Big) = \pm 2\sqrt{3}$

Case 1:

$\Big(x - \frac{1}{x}\Big) = 2\sqrt{3}$

We know that,

$\Rightarrow \Big(x^3 - \frac{1}{x^3}\Big) = \Big(x - \frac{1}{x}\Big)^3 + 3\Big(x - \frac{1}{x}\Big)$

Substituting values we get :

$$\begin{aligned}\Rightarrow \Big(x^3 - \frac{1}{x^3}\Big) = (2\sqrt{3})^3 + 3(2\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \frac{1}{x^3}\Big) = (8 \times 3\sqrt{3}) + (6\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \frac{1}{x^3}\Big) = (24\sqrt{3}) + (6\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \frac{1}{x^3}\Big) = 30\sqrt{3} \\[1em]\end{aligned}$$

Case 1:

$\Big(x - \frac{1}{x}\Big) = -2\sqrt{3}$

We know that,

$\Rightarrow \Big(x^3 - \frac{1}{x^3}\Big) = \Big(x - \frac{1}{x}\Big)^3 + 3\Big(x - \frac{1}{x}\Big)$

Substituting values we get :

$$\begin{aligned}\Rightarrow \Big(x^3 - \frac{1}{x^3}\Big) = (-2\sqrt{3})^3 + 3(-2\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \frac{1}{x^3}\Big) = (-8 \times 3\sqrt{3}) + (-6\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \frac{1}{x^3}\Big) = (-24\sqrt{3}) - (6\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \frac{1}{x^3}\Big) = -30\sqrt{3} \\[1em]\end{aligned}$$

Hence, $\Big(x^3 - \frac{1}{x^3}\Big) = \pm 30\sqrt{3}.$

(i) Given,

$a^2 + \frac{1}{a^2} = 23$

Using identity,

$$\begin{aligned}\Rightarrow \Big(a + \frac{1}{a}\Big)^2 = a^2 + \frac{1}{a^2} + 2 \\[1em] \Rightarrow \Big(a + \frac{1}{a}\Big)^2 = 23 + 2 \\[1em] \Rightarrow \Big(a + \frac{1}{a}\Big)^2 = 25 \\[1em] \Rightarrow \Big(a + \frac{1}{a}\Big) = \sqrt{25} \\[1em] \Rightarrow \Big(a + \frac{1}{a}\Big) = \pm 5\end{aligned}$$

Hence, $\Big(a + \frac{1}{a}\Big) = \pm 5.$

(ii) Given,

$\Big(a + \frac{1}{a}\Big) = \pm 5.$

Case 1:

$\Big(a + \frac{1}{a}\Big) = +5.$

We know that,

$\Rightarrow \Big(a + \frac{1}{a}\Big)^3 = a^3 + \frac{1}{a^3} + 3\Big(a + \frac{1}{a}\Big)$

Substituting values we get :

$$\begin{aligned}\Rightarrow 5^3 = a^3 + \frac{1}{a^3} + 3 \times 5 \\[1em] \Rightarrow 125 = a^3 + \frac{1}{a^3} + 15 \\[1em] \Rightarrow a^3 + \frac{1}{a^3} = 125 - 15 = 110.\end{aligned}$$

Case 2:

$\Big(a + \frac{1}{a}\Big) = -5.$

We know that,

$\Rightarrow \Big(a + \frac{1}{a}\Big)^3 = a^3 + \frac{1}{a^3} + 3\Big(a + \frac{1}{a}\Big)$

Substituting values we get :

$$\begin{aligned}\Rightarrow (-5)^3 = a^3 + \frac{1}{a^3} + 3 \times (-5) \\[1em] \Rightarrow -125 = a^3 + \frac{1}{a^3} - 15 \\[1em] \Rightarrow a^3 + \frac{1}{a^3} = -125 + 15 = -110.\end{aligned}$$

Hence, $a^3 + \frac{1}{a^3} = \pm 110.$

(i) Given,

$a - \frac{1}{a} = \sqrt{5}$

Using identity,

$$\begin{aligned}\Rightarrow \Big(a + \frac{1}{a}\Big)^2 - \Big(a - \frac{1}{a}\Big)^2 = 4 \\[1em] \Rightarrow \Big(a + \frac{1}{a}\Big)^2 - (\sqrt{5})^2 = 4 \\[1em] \Rightarrow \Big(a + \frac{1}{a}\Big)^2 - 5 = 4 \\[1em] \Rightarrow \Big(a + \frac{1}{a}\Big)^2 = 4 + 5 \\[1em] \Rightarrow \Big(a + \frac{1}{a}\Big)^2 = 9 \\[1em] \Rightarrow \Big(a + \frac{1}{a}\Big) = \sqrt{9} \\[1em] \Rightarrow \Big(a + \frac{1}{a}\Big) = \pm 3\end{aligned}$$

Hence, $\Big(a + \frac{1}{a}\Big) = \pm 3.$

(ii) Given,

$\Big(a + \frac{1}{a}\Big) = \pm 3.$

Case 1:

$\Big(a + \frac{1}{a}\Big) = +3.$

We know that,

$\Rightarrow \Big(a + \frac{1}{a}\Big)^3 = a^3 + \frac{1}{a^3} + 3\Big(a + \frac{1}{a}\Big)$

Substituting values we get :

$$\begin{aligned}\Rightarrow 3^3 = a^3 + \frac{1}{a^3} + 3 \times 3 \\[1em] \Rightarrow 27 = a^3 + \frac{1}{a^3} + 9 \\[1em] \Rightarrow a^3 + \frac{1}{a^3} = 27 - 9 = 18.\end{aligned}$$

Case 2:

$\Big(a + \frac{1}{a}\Big) = -3.$

We know that,

$\Rightarrow \Big(a + \frac{1}{a}\Big)^3 = a^3 + \frac{1}{a^3} + 3\Big(a + \frac{1}{a}\Big)$

Substituting values we get :

$$\begin{aligned}\Rightarrow (-3)^3 = a^3 + \frac{1}{a^3} + 3 \times (-3) \\[1em] \Rightarrow -27 = a^3 + \frac{1}{a^3} - 9 \\[1em] \Rightarrow a^3 + \frac{1}{a^3} = -27 + 9 = -18.\end{aligned}$$

Hence, $a^3 + \frac{1}{a^3} = \pm 18.$

(i) Given,

$a^2 + \frac{1}{a^2} = 27$

Using identity,

$$\begin{aligned}\Rightarrow \Big(a - \frac{1}{a}\Big)^2 = a^2 + \frac{1}{a^2} - 2 \\[1em] \Rightarrow \Big(a - \frac{1}{a}\Big)^2 = 27 - 2 \\[1em] \Rightarrow \Big(a - \frac{1}{a}\Big)^2 = 25 \\[1em] \Rightarrow \Big(a - \frac{1}{a}\Big)^2 = \sqrt{25} \\[1em] \Rightarrow \Big(a - \frac{1}{a}\Big)^2 = \pm 5\end{aligned}$$

Hence, $\Big(a - \frac{1}{a}\Big)^2 = \pm 5.$

(ii) Given,

$\Big(a + \frac{1}{a}\Big) = \pm 5.$

Case 1:

$\Big(a + \frac{1}{a}\Big) = +5.$

We know that,

$\Rightarrow \Big(a^3 - \frac{1}{a^3}\Big) = \Big(a - \frac{1}{a}\Big)^3 + 3 \times a \times \frac{1}{a} \Big(a - \frac{1}{a}\Big)$

Substituting values we get :

$$\begin{aligned}\Rightarrow \Big(a^3 - \frac{1}{a^3}\Big) = (5)^3 + 3(5) \\[1em] \Rightarrow \Big(a^3 - \frac{1}{a^3}\Big) = (125) + (15) \\[1em] \Rightarrow \Big(a^3 - \frac{1}{a^3}\Big) = 140.\end{aligned}$$

Case 2:

$\Big(a + \frac{1}{a}\Big) = -5.$

We know that,

$\Rightarrow \Big(a^3 - \frac{1}{a^3}\Big) = \Big(a - \frac{1}{a}\Big)^3 + 3 \times a \times \frac{1}{a} \Big(a - \frac{1}{a}\Big)$

Substituting values we get :

$$\begin{aligned}\Rightarrow \Big(a^3 - \frac{1}{a^3}\Big) = (-5)^3 + 3(-5) \\[1em] \Rightarrow \Big(a^3 - \frac{1}{a^3}\Big) = (-125) + (-15) \\[1em] \Rightarrow \Big(a^3 - \frac{1}{a^3}\Big) = -140.\end{aligned}$$

Hence, $a^3 - \frac{1}{a^3} = \pm 140$

(i) Given,

$$\begin{aligned}\Rightarrow x^2 + \frac{1}{25x^2} = 8\frac{3}{5} \\[1em] \Rightarrow x^2 + \frac{1}{25x^2} = \frac{43}{5}\end{aligned}$$

We know that,

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{5x}\Big)^2 = x^2 + \Big(\frac{1}{5x}\Big)^2 + 2 \times x \times \frac{1}{5x} \\[1em] \Rightarrow \Big(x + \frac{1}{5x}\Big)^2 = x^2 + \frac{1}{25x^2} + \frac{2}{5} \\[1em] \Rightarrow \Big(x + \frac{1}{5x}\Big)^2 = \frac{43}{5} + \frac{2}{5} \\[1em] \Rightarrow \Big(x + \frac{1}{5x}\Big)^2 = \frac{45}{5} \\[1em] \Rightarrow \Big(x + \frac{1}{5x}\Big)^2 = 9 \\[1em] \Rightarrow \Big(x + \frac{1}{5x}\Big) = \sqrt{9} \\[1em] \Rightarrow \Big(x + \frac{1}{5x}\Big) = \pm 3 \\[1em]\end{aligned}$$

Hence, $\Big(x + \frac{1}{5x}\Big) = \pm 3.$

(ii) We know that,

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{5x}\Big)^3 = (x)^3 + \Big(\frac{1}{5x}\Big)^3 + 3 \times x \times \frac{1}{5x} \times \Big(x + \frac{1}{5x}\Big) \\[1em] \Rightarrow \Big(x + \frac{1}{5x}\Big)^3 = (x)^3 + \Big(\frac{1}{5x}\Big)^3 + \frac{3}{5} \times \Big(x + \frac{1}{5x}\Big) \text{ .........(1)}\end{aligned}$$

Given,

$\Big(x + \frac{1}{5x}\Big) = \pm 3.$

Case 1:

$\Big(x + \frac{1}{5x}\Big) = +3.$

Substituting values we get :

$$\begin{aligned}\Rightarrow (3)^3 = x^3 + \frac{1}{125x^3} + \frac{3}{5}\times(3) \\[1em] \Rightarrow 27 = x^3 + \frac{1}{125x^3} + \frac{9}{5} \\[1em] \Rightarrow x^3 + \frac{1}{125x^3} = 27 - \frac{9}{5} \\[1em] \Rightarrow x^3 + \frac{1}{125x^3} = \frac{135-9}{5} \\[1em] \Rightarrow x^3 + \frac{1}{125x^3} = \frac{126}{5} \\[1em] \Rightarrow x^3 + \frac{1}{125x^3} = 25\frac{1}{5}\end{aligned}$$

Case 2:

$\Big(x + \frac{1}{5x}\Big) = -3.$

Substituting values in equation (1), we get :

$$\begin{aligned}\Rightarrow (-3)^3 = x^3 + \frac{1}{125x^3} + \frac{3}{5}\times(-3) \\[1em] \Rightarrow -27 = x^3 + \frac{1}{125x^3} - \frac{9}{5} \\[1em] \Rightarrow x^3 + \frac{1}{125x^3} = -27 + \frac{9}{5} \\[1em] \Rightarrow x^3 + \frac{1}{125x^3} = \frac{-135 + 9}{5} \\[1em] \Rightarrow x^3 + \frac{1}{125x^3} = -\frac{126}{5} \\[1em] \Rightarrow x^3 + \frac{1}{125x^3} = -25\frac{1}{5}\end{aligned}$$

Hence, $x^3 + \frac{1}{25x^3} = \pm 25\frac{1}{5}.$

Given,

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{x}\Big)^2 = 3 \\[1em] \Rightarrow \Big(x + \frac{1}{x}\Big) = \pm \sqrt{3}.\end{aligned}$$

We know that,

$\Rightarrow \Big(x^3 + \frac{1}{x^3}\Big) = \Big(x + \frac{1}{x}\Big)^3 - 3\Big(x + \frac{1}{x}\Big)$

Case 1 :

$\Big(x + \frac{1}{x}\Big) = + \sqrt{3}$

Substituting values we get :

$$\begin{aligned}\Rightarrow \Big(x^3 + \frac{1}{x^3}\Big) = (\sqrt{3})^3 - 3 \times \sqrt{3} \\[1em] \Rightarrow \Big(x^3 + \frac{1}{x^3}\Big) = 3\sqrt{3} - 3\sqrt{3} = 0.\end{aligned}$$

Case 2 :

$\Big(x + \frac{1}{x}\Big) = - \sqrt{3}$

Substituting values we get :

$$\begin{aligned}\Rightarrow \Big(x^3 + \frac{1}{x^3}\Big) = (-\sqrt{3})^3 - 3 \times -\sqrt{3} \\[1em] \Rightarrow \Big(x^3 + \frac{1}{x^3}\Big) = -3\sqrt{3} + 3\sqrt{3} = 0.\end{aligned}$$

Hence, proved that $\Big(x^3 + \frac{1}{x^3}\Big) = 0$.

Given,

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{b}{c} \\[1em] \Rightarrow a = \frac{b}{c} \times b \\[1em] \Rightarrow a \times c = b^2 \\[1em] \Rightarrow ac = b^2\end{aligned}$$

To prove,

(a + b + c)(a - b + c) = a² + b² + c².

Solving L.H.S,

⇒ (a + b + c)(a - b + c)

⇒ a(a - b + c) + b(a - b + c) + c(a - b + c)

⇒ a² - ab + ac + ab - b² + bc + ca - bc + c²

= a² + 2ac - b² + c²

Substituting, ac = b²,

⇒ a² + 2(b²) - b² + c²

⇒ a² + b² + c².

Since, L.H.S. = R.H.S.

Hence, proved that (a + b + c)(a - b + c) = a² + b² + c².

(i) Given,

⇒ (3a + 4b)(9a² - 12ab + 16b²)

⇒ (3a + 4b)(3a)² + (4b)² - 3a × 4b

Using identity,

⇒ (a + b)(a² - ab + b²) = (a³ - b³)

⇒ (3a)³ + (4b)³

Hence, (3a + 4b)(9a² - 12ab + 16b²) = 27a³ + 64b³.

(ii) Using identity,

⇒ (a - b)(a² + ab + b²) = (a³ - b³)

Given,

$$\begin{aligned}\Rightarrow \Big(y - \frac{6}{y}\Big)\Big(y^2 + 6 + \frac{36}{y^2}\Big) \\[1em] \Rightarrow \Big(y - \frac{6}{y}\Big)\Big(y^2 + y \times \frac{6}{y} + \Big(\frac{6}{y}\Big)^2\Big) \\[1em] \Rightarrow y^3 - \Big(\frac{6}{y}\Big)^3 \\[1em] \Rightarrow y^3 - \frac{216}{y^3}.\end{aligned}$$

Hence, $\Big(y - \frac{6}{y}\Big) \Big(y^2 + 6 + \frac{36}{y^2}\Big) = y^3 - \frac{216}{y^3}$.

Taking,

a = 3x, b = -5y and c = -4.

⇒ a² + b² + c² - ab - bc - ca = (3x)² + (-5y)² + (-4)² - 3x × (-5y) - (-5y) × (-4) - (-4) × 3x

⇒ a² + b² + c² - ab - bc - ca = 9x² + 25y² + 16 + 15xy - 20y + 12x.

Using identity,

(a + b + c)(a² + b² + c² - ab - bc - ca) = a³ + b³ + c³ - 3abc

⇒ (3x - 5y - 4)(9x² + 25y² + 16 + 15xy - 20y + 12x) = (3x)³ + (-5y)³ + (-4)³ - 3 × 3x × (-5y) × (-4)

⇒ (3x - 5y - 4)(9x² + 25y² + 16 + 15xy - 20y + 12x) = 27x³ - 125y³ - 64 - 180xy.

Hence, (3x - 5y - 4)(9x² + 25y² + 16 + 15xy - 20y + 12x) = 27x³ - 125y³ - 64 - 180xy.

(i) Given,

⇒ (x + 6)(x + 2)

⇒ (x² + 2x + 6x + 12)

⇒ (x² + 8x + 12).

Hence, (x + 6)(x + 2) = (x² + 8x + 12).

Given,

⇒ (x + 8)(x - 3)

⇒ (x² - 3x + 8x - 24)

⇒ (x² + 5x - 24).

Hence, (x + 8)(x - 3) = (x² + 5x - 24).

Given,

⇒ (x - 5)(x - 7)

⇒ (x² - 7x - 5x + 35)

⇒ (x² - 12x + 35).

Hence, (x - 5)(x - 7) = (x² - 12x + 35).

Given,

⇒ (2 - x)(4 - x)

⇒ (8 - 2x - 4x + x²)

⇒ (8 - 6x + x²).

Hence, (2 - x)(4 - x) = (8 - 6x + x²).

Given,

⇒ (y - 7)(y + 4)

⇒ (y² + 4y - 7y - 28)

⇒ (y² - 3y - 28).

Hence, (y - 7)(y + 4) = (y² - 3y - 28).

Given,

⇒ (ab + 3)(ab - 2)

⇒ (a²b² - 2ab + 3ab - 6)

⇒ (a²b² + ab - 6).

Hence, (ab + 3)(ab - 2) = (a²b² + ab - 6).

Given,

⇒ (5 - xy)(2 + xy)

⇒ (10 + 5xy - 2xy - x²y²)

⇒ (10 + 3xy - x²y²).

Hence, (5 - xy)(2 + xy) = (10 + 3xy - x²y²).

Given,

⇒ (x² + 1)(x² + 2)

⇒ (x⁴ + 2x² + x² + 2)

⇒ (x⁴ + 3x² + 2).

Hence, (x² + 1)(x² + 2) = (x⁴ + 3x² + 2).

Given,

⇒ (3 - x²)(5 + x²)

⇒ (15 + 3x² - 5x² - x⁴)

⇒ (15 - 2x² - x⁴).

Hence, (3 - x²)(5 + x²) = (15 - 2x² - x⁴).

Given,

⇒ (6 - x)(x + 5)

⇒ (6x + 30 - x² - 5x)

⇒ (30 - x² + x).

Hence, (6 - x)(x + 5) = (30 - x² + x).

Given,

⇒ (2x + 3)(3x + 5)

⇒ (6x² + 10x + 9x + 15)

⇒ (6x² + 19x + 15).

Hence, (2x + 3)(3x + 5) = (6x² + 19x + 15).

Given,

⇒ (7m - 2)(4m + 3)

⇒ (28m² + 21m - 8m - 6)

⇒ (28m² + 13m - 6).

Hence, (7m - 2)(4m + 3) = (28m² + 13m - 6).

Given,

⇒ (2y - 3)(3y - 5)

⇒ (6y² - 10y - 9y + 15)

⇒ (6y² - 19y + 15).

Hence, (2y - 3)(3y - 5) = (6y² - 19y + 15).

Given,

⇒ (3a² - b²)(2a² + 5b²)

⇒ (6a⁴ + 15a²b² - 2a²b² - 5b⁴)

⇒ (6a⁴ + 13a²b² - 5b⁴).

Hence, (3a² - b²)(2a² + 5b²) = 6a⁴ + 13a²b² - 5b⁴.

Given,

a + b = 12 and ab = 35

Using identity,

⇒ (a + b)² - (a - b)² = 4ab

⇒ (a − b)² = (a + b)² - 4ab

⇒ (a − b)² = (12)² - 4(35)

⇒ (a − b)² = 144 - 140

⇒ (a − b)² = 4

Hence, Option 2 is the correct option.

Given,

a + b = 16 and a - b = 2

Using identity,

⇒ (a + b)² - (a - b)² = 4ab

⇒ 16² - 2² = 4ab

⇒ 4ab = 256 - 4

⇒ 4ab = 252

⇒ ab = $\frac{252}{4}$

⇒ ab = 63.

Hence, Option 3 is the correct option.

Given,

$p + \frac{1}{p} = \frac{5}{2}$

Using identity,

$$\begin{aligned}\Rightarrow \Big(p + \frac{1}{p}\Big)^2 = p^2 + \frac{1}{p^2} + 2 \\[1em] \Rightarrow \Big(\frac{5}{2}\Big)^2 = \Big(p^2 + \frac{1}{p^2}\Big) + 2 \\[1em] \Rightarrow \frac{25}{4} = \Big(p^2 + \frac{1}{p^2}\Big) + 2 \\[1em] \Rightarrow \frac{25}{4} - 2 = \Big(p^2 + \frac{1}{p^2}\Big) \\[1em] \Rightarrow \Big(\frac{25 - 8}{4}\Big) = \Big(p^2 + \frac{1}{p^2}\Big) \\[1em] \Rightarrow \Big(p^2 + \frac{1}{p^2}\Big) = \frac{17}{4} \\[1em]\end{aligned}$$

Hence, Option 4 is the correct option.

Given,

$x + \frac{1}{x} = 5$

Using identity,

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{x}\Big)^2 = x^2 + \frac{1}{x^2} + 2 \times x \times \frac{1}{x}\\[1em] \Rightarrow (5)^2 = x^2 + \frac{1}{x^2} + 2 \\[1em] \Rightarrow 25 - 2 = x^2 + \frac{1}{x^2} \\[1em] \Rightarrow x^2 + \frac{1}{x^2} = 23\\[1em]\end{aligned}$$

We know that,

$$\begin{aligned}\Rightarrow \Big(x - \frac{1}{x}\Big)^2 = x^2 + \frac{1}{x^2} - 2 \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big)^2 = 23 - 2 \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big)^2 = 21 \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big) = \pm \sqrt{21} \\[1em]\end{aligned}$$

Hence, Option 1 is the correct option.

Given,

$y + \frac{1}{y} = 2$

$$\begin{aligned}\Rightarrow \frac{y^2 + 1}{y} = 2 \\[1em] \Rightarrow y^2 + 1 = 2y \\[1em] \Rightarrow y^2 - 2y + 1 = 0 \\[1em] \Rightarrow (y - 1)^2 = 0 \\[1em] \Rightarrow y-1 = 0 \\[1em] \Rightarrow y = 1\end{aligned}$$

Substituting value of y, we get :

$\Rightarrow y^5 + \frac{1}{y^5} = (1)^5 + \frac{1}{(1)^5} = 1 + 1 = 2$.

Hence, option 3 is correct option.

Given,

$a^2 - \frac{1}{a^2} = 7$

Upon squaring both sides,

$$\begin{aligned}\Rightarrow \Big(a^2 - \frac{1}{a^2}\Big)^2 = a^4 + \frac{1}{a^4} - 2 \\[1em] \Rightarrow (7)^2 = \Big(a^4 + \frac{1}{a^4}\Big) - 2 \\[1em] \Rightarrow 49 = \Big(a^4 + \frac{1}{a^4}\Big) - 2 \\[1em] \Rightarrow \Big(a^4 + \frac{1}{a^4}\Big) = 49 + 2 \\[1em] \Rightarrow \Big(a^4 + \frac{1}{a^4}\Big) = 51 \\[1em]\end{aligned}$$

Hence, Option 2 is the correct option.

Given,

x + y + z = 0

Using identity,

⇒ x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)

⇒ x³ + y³ + z³ - 3xyz = (0) × (x² + y² + z² - xy - yz - zx)

⇒ x³ + y³ + z³ - 3xyz = 0

⇒ x³ + y³ + z³ = 3xyz.

Hence, Option 2 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \frac{a}{b} + \frac{b}{a} = 1 \\[1em] \Rightarrow \frac{a^2 + b^2}{ab} = 1 \\[1em] \Rightarrow a^2 + b^2 = ab \\[1em] \Rightarrow a^2 + b^2 - ab = 0.\end{aligned}$$

We know that,

⇒ a³ + b³ = (a + b)(a² - ab + b²)

⇒ a³ + b³ = (a + b)(0)

⇒ a³ + b³ = 0.

Hence, Option 1 is the correct option.

Given,

Upon squaring $p + \frac{1}{p} = x$ we get,

$$\begin{aligned}\Rightarrow \Big(p + \frac{1}{p}\Big)^2 = x^2 \\[1em] \Rightarrow \Big(p^2 + \frac{1}{p^2} + 2\Big) = x^2 \text{ ....(1)}\end{aligned}$$

Upon squaring $p - \frac{1}{p} = y$ we get,

$$\begin{aligned}\Rightarrow \Big(p - \frac{1}{p}\Big)^2 = y^2 \\[1em] \Rightarrow \Big(p^2 + \frac{1}{p^2} - 2\Big) = y^2 \text{ ....(2)}\end{aligned}$$

Subtracting (2) from (1) we get,

$$\begin{aligned}\Rightarrow \Big(p^2 + \frac{1}{p^2} + 2\Big) - \Big(p^2 + \frac{1}{p^2} - 2\Big) = x^2 - y^2 \\[1em] \Rightarrow \Big(p^2 + \frac{1}{p^2} + 2 - p^2 - \frac{1}{p^2} + 2\Big) = x^2 - y^2 \\[1em] \Rightarrow x^2 - y^2 = 4.\end{aligned}$$

Hence, Option 3 is the correct option.

Given,

$$\begin{aligned}\Rightarrow a = \frac{1}{a - 7} \\[1em] \Rightarrow a \times (a-7) = 1 \\[1em] \Rightarrow a - 7 = \frac{1}{a} \\[1em] \Rightarrow a - \frac{1}{a} = 7\end{aligned}$$

Hence, option 3 is correct option.

Given,

⇒ 2.51 × 2.51 + 1.31 × 1.31 − 2.62 × 2.51

⇒ 2.51² + 1.31² - 2(1.31)(2.51)

⇒ 2.51² + 1.31² - 2(2.51)(1.31)

⇒ (2.51 - 1.31)²     [(a - b)² = a² + b² - 2ab]

⇒ (1.2)²

⇒ 1.44

Hence, Option 1 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{x}\Big)^2 = 3 \\[1em] \Rightarrow \Big(x + \frac{1}{x}\Big) = \pm \sqrt{3}\end{aligned}$$

Case 1:

$\Rightarrow \Big(x + \frac{1}{x}\Big) = \sqrt{3}$

Using identity,

$$\begin{aligned}\Rightarrow \Big(x^3 + \frac{1}{x^3}\Big) = \Big(x + \frac{1}{x}\Big)^3 - 3\Big(x + \frac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^3 + \frac{1}{x^3}\Big) = (\sqrt{3})^3 - 3(\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 + \frac{1}{x^3}\Big) = (3\sqrt{3}) - 3(\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 + \frac{1}{x^3}\Big) = 0\end{aligned}$$

Case 2:

$\Rightarrow \Big(x + \frac{1}{x}\Big) = -\sqrt{3}$

Using identity,

$$\begin{aligned}\Rightarrow \Big(x^3 + \frac{1}{x^3}\Big) = \Big(x + \frac{1}{x}\Big)^3 - 3\Big(x + \frac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^3 + \frac{1}{x^3}\Big) = (-\sqrt{3})^3 - 3(-\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 + \frac{1}{x^3}\Big) = (-3\sqrt{3}) + 3(\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 + \frac{1}{x^3}\Big) = 0\end{aligned}$$

Hence, Option 4 is the correct option.

Given,

3a + 5b = 15 and 9a² + 25b² = 75

⇒ (3a + 5b)² = (3a)² + (5b)² + 2 × (3a) × (5b)

⇒ (3a + 5b)² = 9a² + 25b² + 30ab

⇒ 15² = 75 + 30ab

⇒ 225 = 75 + 30ab

⇒ 30ab = 225 - 75

⇒ 30ab = 150

⇒ ab = $\frac{150}{30}$

⇒ ab = 5.

Hence, Option 2 is the correct option.

Given,

l + m − n = 9

l² + m² + n² = 31

Solving,

⇒ [(l + m) − (n)]² = (l + m)² + n² - 2 × (l + m) × n

⇒ [(l + m) − (n)]² = l² + m² + 2 × l × m + n² - 2ln - 2mn

⇒ (9)² = l² + m² + 2lm + n² - 2ln - 2mn

⇒ 81 = l² + m² + n² + 2lm - 2ln - 2mn

⇒ 81 = 31 + 2(lm - ln - mn)

⇒ 2(lm - ln - mn) = 81 - 31

⇒ (lm - ln - mn) = $\frac{50}{2}$

⇒ (lm - ln - mn) = 25

⇒ -(lm - ln - mn) = -25

⇒ (mn + nl - lm) = -25

Hence, Option 1 is the correct option.

Given,

a - b + c = 6

a² + b² + c² = 38

We know that,

⇒ [(a - b) + (c)]² = (a - b)² + c² + 2 × (a - b) × c

⇒ [(a - b) + (c)]² = a² + b² - 2 × a × b + c² + 2ac - 2bc

⇒ (6)² = a² + b² + c² - 2(ab - ac + bc)

⇒ 36 = 38 - 2(ab - ac + bc)

⇒ 2(ab + bc − ca) = 38 - 36

⇒ 2(ab + bc − ca) = 2

⇒ (ab + bc − ca) = $\frac{2}{2}$

⇒ (ab + bc − ca) = 1.

Hence, Option 2 is the correct option.

We know that,

⇒ x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)

If x + y + z = 0, then :

⇒ x³ + y³ + z³ - 3xyz = 0

⇒ x³ + y³ + z³ = 3xyz.

So, reason (R) is true.

⇒ 26 + (-15) + (-11)

⇒ 26 - 26

⇒ 0

Since, 26 + (-15) + (-11) = 0,

∴ (26)³ + (−15)³ + (−11)³ = 3 × 26 × -15 × -11

Assertion (A) is false.

Thus, A is false and R is true.

Hence, Option 2 is the correct option.

Given,

$$\begin{aligned}\Rightarrow x^2 - 2x - 1 = 0 \\[1em] \Rightarrow x^2 - 1 = 2x \\[1em] \Rightarrow \frac{x^2 - 1}{x} = 2 \\[1em] \Rightarrow \frac{x^2}{x} - \frac{1}{x} = 2 \\[1em] \Rightarrow x - \frac{1}{x} = 2 \\[1em]\end{aligned}$$

Using identity,

$\Big(x - \frac{1}{x}\Big)^2 = x^2 + \frac{1}{x^2} - 2$

Substituting,

$$\begin{aligned}\Rightarrow (2)^2 = x^2 + \frac{1}{x^2} - 2 \\[1em] \Rightarrow 4 = x^2 + \frac{1}{x^2} - 2 \\[1em] \Rightarrow x^2 + \frac{1}{x^2} = 4 + 2 \\[1em] \Rightarrow x^2 + \frac{1}{x^2} = 6.\end{aligned}$$

Assertion (A) is true.

⇒ (x - 1)² = x² + 1² - 2(x)(1)

⇒ (x - 1)² = x² - 2x + 1

Reason (R) is false.

A is true, R is false

Hence, Option 1 is the correct option.

Using identity,

(a - b)³ = a³ - b³ - 3ab(a - b)

So, reason (R) is true.

⇒ (1 - 3x)³ = 1³ - (3x)³ - 3 × 1 × 3x (1 - 3x)

⇒ (1 - 3x)³ = 1 - 27x³ - 9x(1 - 3x)

⇒ (1 - 3x)³ = 1 - 27x³ - 9x + 27x²

So, assertion (A) is false.

Hence, Option 2 is the correct option.

Given,

(−28)³ + (18)³ + (10)³

We know that,

If a + b + c = 0, then a³ + b³ + c³ = 3abc.

Since, -28 + 18 + 10 = 0

⇒ (−28)³ + 18³ + 10³ = 3 × (-28) × 18 × 10

⇒ (−28)³ + 18³ + 10³ = -15120.

Hence, Option 2 is the correct option.

Given,

x + y = -4

We know that,

⇒ x³ + y³ = (x + y)³ - 3xy(x + y)

⇒ x³ + y³ = (-4)³ - 3xy(-4)

⇒ x³ + y³ = -64 + 12xy

⇒ x³ + y³ - 12xy + 64 = 0.

Hence, Option 4 is the correct option.

Given,

a + b + c = 0

⇒ a + b = -c

⇒ b + c = -a

⇒ c + a = -b

Substituting the above values in $\frac{(a + b)^2}{ab} + \frac{(b + c)^2}{bc} + \frac{(c + a)^2}{ac}$, we get :

$$\begin{aligned}\Rightarrow \frac{(-c)^2}{ab} + \frac{(-a)^2}{bc} + \frac{(-b)^2}{ca} \\[1em] \Rightarrow \frac{c^2}{ab} + \frac{a^2}{bc} + \frac{b^2}{ca} \\[1em] \Rightarrow \frac{a^3 + b^3 + c^3}{abc} \text{ ......(1)}\end{aligned}$$

We know that,

If, a + b + c = 0 then a³ + b³ + c³ = 3abc

Substituting the value of a³ + b³ + c³ in (1), we get :

$$\begin{aligned}\Rightarrow \frac{3abc}{abc} \\[1em] \Rightarrow 3.\end{aligned}$$

Hence, option 3 is correct option.

Given,

$$\begin{aligned}\Rightarrow \frac{a}{b} + \frac{b}{a} = -1 \\[1em] \Rightarrow \frac{a^2 + b^2}{ab} = -1 \\[1em] \Rightarrow a^2 + b^2 = -ab \\[1em] \Rightarrow a^2 + b^2 + ab = 0\end{aligned}$$

Using identity,

⇒ a³ - b³ = (a - b)(a² + b² + ab)

⇒ a³ - b³ = (a - b)(0)

⇒ a³ - b³ = 0.

Hence, option 4 is correct option.

Given,

$x^4 + \frac{1}{x^4} = 119$, and x > 1.

$x^2 + \frac{1}{x^2}$

Using identity

$\Big(x^2 + \frac{1}{x^2}\Big)^2 = x^4 + \frac{1}{x^4} + 2$

Substitute the given value:

$$\begin{aligned}\Rightarrow \Big(x^2 + \frac{1}{x^2}\Big)^2 = 119 + 2 \\[1em] \Rightarrow \Big(x^2 + \frac{1}{x^2}\Big)^2 = 121 \\[1em] \Rightarrow x^2 + \frac{1}{x^2} = \sqrt{121} \\[1em] \Rightarrow x^2 + \frac{1}{x^2} = \pm 11\end{aligned}$$

Since $x \gt 1, x^2 + \frac{1}{x^2}$ must be positive.

So, $x^2 + \frac{1}{x^2} = 11$

$x - \frac{1}{x}$

Using identity

$\Big(x - \frac{1}{x}\Big)^2 = x^2 + \frac{1}{x^2} - 2$

Substitute the value:

$$\begin{aligned}\Rightarrow \Big(x - \frac{1}{x}\Big)^2 = 11 - 2 \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big)^2 = 9 \\[1em] \Rightarrow x - \frac{1}{x} = \pm\sqrt{9} \\[1em] \Rightarrow x - \frac{1}{x} = \pm 3\end{aligned}$$

Since $x \gt 1, x$ is greater than $\frac{1}{x}$, so $x - \frac{1}{x}$ must be positive.

So, $x - \frac{1}{x} = 3$

$x^3 - \frac{1}{x^3}$

By using the identity:

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Let $a = x$ and $b = \frac{1}{x}$.

$$\begin{aligned}\Rightarrow x^3 - \frac{1}{x^3} = \Big(x - \frac{1}{x}\Big)\Big(x^2 + x \cdot \frac{1}{x} + \frac{1}{x^2}\Big) \\[1em] \Rightarrow x^3 - \frac{1}{x^3} = \Big(x - \frac{1}{x}\Big)\Big(x^2 + \frac{1}{x^2} + 1\Big)\\[1em] \Rightarrow x^3 - \frac{1}{x^3} = (3)(11 + 1) \\[1em] \Rightarrow x^3 - \frac{1}{x^3} = (3)(12) \\[1em] \Rightarrow x^3 - \frac{1}{x^3} = 36\end{aligned}$$

Hence, $x^3 - \frac{1}{x^3} = 36$.

Given,

$\frac{1}{x + y} = \frac{1}{x} + \frac{1}{y}$

$$\begin{aligned}\Rightarrow \frac{1}{x + y} = \frac{y + x}{xy} \\[1em] \Rightarrow xy = (x + y)(x + y) \\[1em] \Rightarrow xy = x^2 + y^2 + 2xy \\[1em] \Rightarrow 0 = x^2 + y^2 + 2xy - xy \\[1em] \Rightarrow x^2 + xy + y^2 = 0\end{aligned}$$

Using identity,

x³ - y³ = (x - y)(x² + xy +y²)

x³ - y³ = (x - y)(0)

x³ - y³ = 0

Hence, x³ - y³ = 0.

Given,

$$\begin{aligned}\Rightarrow x^2 + y^2 + \frac{1}{x^2} + \frac{1}{y^2} = 4 \\[1em] \Rightarrow x^2 + y^2 + \frac{1}{x^2} + \frac{1}{y^2} - 4 = 0 \\[1em] \Rightarrow \Big(x^2 + \frac{1}{x^2} - 2\Big) + \Big(y^2 + \frac{1}{y^2} - 2\Big) = 0\end{aligned}$$

Using identity,

$a^2 + \frac{1}{a^2} - 2 = \Big(a - \frac{1}{a}\Big)^2$.

$\Big(x - \frac{1}{x}\Big)^2 + \Big(y - \frac{1}{y}\Big)^2 = 0$

Hence,

$\Big(x - \frac{1}{x}\Big)^2 = 0 \text{ and } \Big(y - \frac{1}{y}\Big)^2 = 0$

$x - \frac{1}{x} = 0 \text{ and } y - \frac{1}{y} = 0$

Solving for x: $$\begin{aligned}\Rightarrow x = \frac{1}{x} \\[1em] \Rightarrow x^2 = 1\end{aligned}$$

Solving for y: $$\begin{aligned}\Rightarrow y = \frac{1}{y} \\[1em] \Rightarrow y^2 = 1\end{aligned}$$

Solving for x² + y²:

x² + y² = 1 + 1

x² + y² = 2

Hence, x² + y² = 2.