Expansions

Given,

⇒ x² + 2xy + y² - 3

⇒ (x + y)² - 3

Substituting values we get :

⇒ (9 + 4)² - 3

⇒ 13² - 3

⇒ 169 - 3

⇒ 166.

Hence, Option 3 is the correct option.

Given,

⇒ x² + y² - 2xy + 7

⇒ (x - y)² + 7

Substituting values we get :

⇒ (5 - 3)² + 7

⇒ 2² + 7

⇒ 4 + 7

⇒ 11.

Hence, Option 1 is the correct option.

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

Given,

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

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

Adding 5 on both sides,

$$\begin{aligned}x^2 + \frac{1}{x^2} + 5 = 2 + 5 \\[1em] \Rightarrow x^2 + \frac{1}{x^2} + 5 = 7\end{aligned}$$

Hence, Option 4 is the correct option.

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

Subtracting 8 on both sides,

$$\begin{aligned}\Rightarrow x^2 + \frac{1}{x^2} - 8 = \Big(x - \frac{1}{x}\Big)^2 + 2 - 8 \\[1em] \Rightarrow x^2 + \frac{1}{x^2} - 8 = 8^2 - 6 \\[1em] \Rightarrow x^2 + \frac{1}{x^2} - 8 = 64 - 6 \\[1em] \Rightarrow x^2 + \frac{1}{x^2} - 8 = 58.\end{aligned}$$

Hence, Option 2 is the correct option.

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

Adding 5 on both sides,

$$\begin{aligned}x^4 + \frac{1}{x^4} + 5 = \Big(x^2 + \frac{1}{x^2}\Big)^2 - 2 + 5\\[1em] \Rightarrow x^4 + \frac{1}{x^4} + 5 = 9^2 + 3 \\[1em] \Rightarrow x^4 + \frac{1}{x^4} + 5 = 84.\end{aligned}$$

Hence, Option 3 is the correct option.

Given,

⇒ x² - 3x + 1 = 0

⇒ x² + 1 = 3x

Dividing above equation by x, we get :

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

Squaring both sides we get :

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

Hence, Option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{7}{8}x + \frac{4}{5}y\Big)^2 \\[1em] \Rightarrow \Big(\frac{7}{8}x\Big)^2 + \Big(\frac{4}{5}y\Big)^2 + 2 \times \frac{7}{8}x \times \frac{4}{5}y \\[1em] \Rightarrow \frac{49}{64}x^2 + \frac{16}{25}y^2 + \frac{7}{5}xy.\end{aligned}$$

Hence, $\Big(\frac{7}{8}x + \frac{4}{5}y\Big)^2 = \frac{49}{64}x^2 + \frac{16}{25}y^2 + \frac{7}{5}xy.$

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{2x}{7} - \frac{7y}{4}\Big)^2 \\[1em] \Rightarrow \Big(\frac{2x}{7}\Big)^2 + \Big(\frac{7y}{4}\Big)^2 - 2 \times \frac{2x}{7} \times \frac{7y}{4} \\[1em] \Rightarrow \frac{4x^2}{49} + \frac{49y^2}{16} - xy.\end{aligned}$$

Hence, $\Big(\frac{2x}{7} - \frac{7y}{4}\Big)^2 = \frac{4x^2}{49} + \frac{49y^2}{16} - xy.$

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{a}{2b} + \frac{2b}{a}\Big)^2 - \Big(\frac{a}{2b} - \frac{2b}{a}\Big)^2 - 4 \\[1em] \Rightarrow \Big(\frac{a}{2b}\Big)^2 + \Big(\frac{2b}{a}\Big)^2 + 2 \times \frac{a}{2b} \times \frac{2b}{a} - \Big[\Big(\frac{a}{2b}\Big)^2 + \Big(\frac{2b}{a}\Big)^2 - 2 \times \frac{a}{2b} \times \frac{2b}{a}\Big] - 4 \\[1em] \Rightarrow \frac{a^2}{4b^2} + \frac{4b^2}{a^2} + 2 - \Big[\frac{a^2}{4b^2} + \frac{4b^2}{a^2} - 2\Big] - 4 \\[1em] \Rightarrow \frac{a^2}{4b^2} + \frac{4b^2}{a^2} + 2 - \frac{a^2}{4b^2} - \frac{4b^2}{a^2} + 2 - 4 \\[1em] \Rightarrow 4 - 4 \\[1em] \Rightarrow 0.\end{aligned}$$

Hence, $\Big(\frac{a}{2b} + \frac{2b}{a}\Big)^2 - \Big(\frac{a}{2b} - \frac{2b}{a}\Big)^2 - 4$ = 0.

(i) By formula,

(x - y)² = (x + y)² - 4xy

$$\begin{aligned}\Rightarrow (x - y)^2 = \Big(\frac{7}{2}\Big)^2 - 4 \times \frac{5}{2} \\[1em] \Rightarrow (x - y)^2 = \frac{49}{4} - 10 \\[1em] \Rightarrow (x - y)^2 = \frac{49 - 40}{4} \\[1em] \Rightarrow (x - y)^2 = \frac{9}{4} \\[1em] \Rightarrow (x - y) = \sqrt{\frac{9}{4}} \\[1em] \Rightarrow x - y = \pm\frac{3}{2}.\end{aligned}$$

Hence, x - y = $\pm \frac{3}{2}$.

(ii) Solving,

$$\begin{aligned}\Rightarrow x^2 - y^2 \\[1em] \Rightarrow (x - y)(x + y) \\[1em] \Rightarrow \pm \frac{3}{2} \times \frac{7}{2} \\[1em] \Rightarrow \pm \frac{21}{4}.\end{aligned}$$

Hence, x² - y² = $\pm \frac{21}{4}$.

(i) By formula,

⇒ (a + b)² = (a - b)² + 4ab

⇒ (a + b)² = (0.9)² + 4 × 0.36

⇒ (a + b)² = 0.81 + 1.44

⇒ (a + b)² = 2.25

⇒ a + b = $\sqrt{2.25} = \pm1.5$

Hence, a + b = $\pm 1.5$

(ii) Solving,

⇒ a² - b²

⇒ (a - b)(a + b)

⇒ 0.9 × $\pm 1.5$

⇒ $\pm 1.35$

Hence, a² - b² = $\pm 1.35$

(i) We know that,

(a - b)² = a² + b² - 2ab .............(1)

(a + b)² = a² + b² + 2ab ..............(2)

Adding equation (1) and (2), we get :

(a - b)² + (a + b)² = 2(a² + b²)

(a² + b²) = $\frac{(a - b)^2 + (a + b)^2}{2}$

Substituting values we get :

$$\begin{aligned}\Rightarrow a^2 + b^2 = \frac{4^2 + 6^2}{2} \\[1em] \Rightarrow a^2 + b^2 = \frac{16 + 36}{2} \\[1em] \Rightarrow a^2 + b^2 = \frac{52}{2} \\[1em] \Rightarrow a^2 + b^2 = 26.\end{aligned}$$

Hence, a² + b² = 26.

(ii) We know that,

(a - b)² = a² + b² - 2ab .............(1)

(a + b)² = a² + b² + 2ab ..............(2)

Subtracting equation (1) from (2), we get :

⇒ (a + b)² - (a - b)² = a² + b² + 2ab - (a² + b² - 2ab)

⇒ (a + b)² - (a - b)² = 4ab

⇒ ab = $\frac{(a + b)^2 - (a - b)^2}{4}$

Substituting values we get :

$$\begin{aligned}\Rightarrow ab = \frac{6^2 - 4^2}{4} \\[1em] = \frac{36 - 16}{4} \\[1em] =\frac{20}{4} \\[1em] = 5.\end{aligned}$$

Hence, ab = 5.

(i) By formula,

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

Hence, $a - \frac{1}{a} = \pm 4\sqrt{2}$.

(ii) Solving,

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

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

(i) By formula,

$\Rightarrow \Big(a + \frac{1}{a}\Big)^2 - \Big(a - \frac{1}{a}\Big)^2$ = 4

Substituting values we get :

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

Hence, $a + \frac{1}{a} = \pm 2\sqrt{17}.$

(ii) By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow a^2 - \frac{1}{a^2} = \pm 2\sqrt{17} \times 8 \\[1em] = \pm 16\sqrt{17}.\end{aligned}$$

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

(i) Given,

⇒ a² - 3a + 1 = 0

⇒ a² + 1 = 3a

Dividing above equation by a, we get :

$\Rightarrow \frac{a^2 + 1}{a} = \frac{3a}{a}$

$\Rightarrow a + \frac{1}{a} = 3$

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

(ii) By formula,

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

Hence, $a^2 + \frac{1}{a^2} = 7$.

(i) Given,

⇒ a² - 5a - 1 = 0

⇒ a² - 1 = 5a

Dividing above equation by a, we get :

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

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

Hence, $a - \frac{1}{a} = 5$.

(ii) By formula,

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

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

(iii) By formula,

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

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

Solving,

⇒ (3x + 4y)² = (3x)² + (4y)² + 2 × 3x × 4y

⇒ (3x + 4y)² = 9x² + 16y² + 24xy

Substituting values we get :

⇒ 16² = 9x² + 16y² + 24 × 4

⇒ 256 = 9x² + 16y² + 96

⇒ 9x² + 16y² = 256 - 96 = 160.

Hence, 9x² + 16y² = 160.

Let two positive numbers be x and y.

Given,

Difference between two positive numbers is 5.

∴ x - y = 5

⇒ x = y + 5 ...........(1)

Given,

Sum of squares of numbers is 73.

∴ x² + y² = 73

Substituting value of x from equation (1) in above equation, we get :

⇒ (y + 5)² + y² = 73

⇒ y² + 5² + 2 × y × 5 + y² = 73

⇒ 2y² + 25 + 10y = 73

⇒ 2y² + 10y + 25 - 73 = 0

⇒ 2y² + 10y - 48 = 0

⇒ 2(y² + 5y - 24) = 0

⇒ y² + 5y - 24 = 0

⇒ y² + 8y - 3y - 24 = 0

⇒ y(y + 8) - 3(y + 8) = 0

⇒ (y - 3)(y + 8) = 0

⇒ y - 3 = 0 or y + 8 = 0

⇒ y = 3 or y = -8.

If y = 3, x = y + 5 = 3 + 5 = 8, xy = 24,

If y = -8, x = y + 5 = -8 + 5 = -3, xy = 24.

Hence, product of numbers = 24.

Given,

⇒ x² - 2x + 1 = 0

⇒ x² + 1 = 2x

Dividing above equation by x, we get :

⇒ $\frac{x^2 + 1}{x} = \frac{2x}{x}$

⇒ $x + \frac{1}{x}$ = 2.

Squaring both sides we get :

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

Squaring both sides we get :

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

Hence, Option 4 is the correct option.

By property,

If a + b + c = 0, then

a³ + b³ + c³ = 3abc

Since, 5 + (-2) + (-3) = 5 - 2 - 3 = 0.

∴ a³ + b³ + c³ = 3 × 5 × -2 × -3 = 90.

Hence, Option 2 is the correct option.

By formula,

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

Substituting values we get :

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

Hence, Option 1 is the correct option.

By property,

If a + b + c = 0, then

a³ + b³ + c³ = 3abc ....(1)

Comparing equation x + 3y + 2z = 0, with a + b + c = 0, we get :

a = x, b = 3y and c = 2z.

Substituting values in equation (1), we get :

x³ + (3y)³ + (2z)³ = 3 × x × 3y × 2z = 18xyz.

Hence, Option 3 is the correct option.

Solving,

(3a - 2b)³ = (3a)³ - (2b)³ - 3 × 3a × 2b(3a - 2b)

= 27a³ - 8b³ - 18ab(3a - 2b)

= 27a³ - 8b³ - 54a²b + 36ab².

Hence, cube of 3a - 2b = 27a³ - 8b³ - 54a²b + 36ab².

Solving,

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

= 125a³ + 27b³ + 45ab(5a + 3b)

= 125a³ + 27b³ + 225a²b + 135ab².

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

Solving,

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

Hence, $\Big(2a + \frac{1}{2a}\Big)^3 = 8a^3 + \frac{1}{8a^3} + 6a + \frac{3}{2a}$.

Solving,

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

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

(i) By formula,

$$\begin{aligned}\Rightarrow \Big(a + \frac{1}{a}\Big)^2 = a^2 + \frac{1}{a^2} + 2 \\[1em] \Rightarrow \Big(a + \frac{1}{a}\Big)^2 = 47 + 2 \\[1em] \Rightarrow \Big(a + \frac{1}{a}\Big)^2 = 49 \\[1em] \Rightarrow a + \frac{1}{a} = \sqrt{49} \\[1em] \Rightarrow a + \frac{1}{a} = \pm 7.\end{aligned}$$

Hence, $a + \frac{1}{a} = \pm 7.$

(ii) By formula,

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

Substituting $a + \frac{1}{a} = 7$ in equation (1), we get :

$$\begin{aligned}\Rightarrow 7^3 = a^3 + \frac{1}{a^3} + 3 \times 7 \\[1em] \Rightarrow 343 = a^3 + \frac{1}{a^3} + 21 \\[1em] \Rightarrow a^3 + \frac{1}{a^3} = 343 - 21 \\[1em] \Rightarrow a^3 + \frac{1}{a^3} = 322.\end{aligned}$$

Substituting $a + \frac{1}{a} = -7$ in equation (1), we get :

$$\begin{aligned}\Rightarrow (-7)^3 = a^3 + \frac{1}{a^3} + 3 \times -7 \\[1em] \Rightarrow -343 = a^3 + \frac{1}{a^3} - 21 \\[1em] \Rightarrow a^3 + \frac{1}{a^3} = -343 + 21 \\[1em] \Rightarrow a^3 + \frac{1}{a^3} = -322.\end{aligned}$$

Hence, $a^3 + \frac{1}{a^3} = \pm 322.$

(i) By formula,

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

Hence, $a - \frac{1}{a} = \pm 4$.

(ii) By formula,

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

Substituting $a - \frac{1}{a} = 4$ we get :

$$\begin{aligned}\Rightarrow 4^3 = a^3 - \frac{1}{a^3} - 3 \times 4 \\[1em] \Rightarrow 64 = a^3 - \frac{1}{a^3} - 12 \\[1em] \Rightarrow 64 + 12 = a^3 - \frac{1}{a^3} \\[1em] \Rightarrow a^3 - \frac{1}{a^3} = 76.\end{aligned}$$

Substituting $a - \frac{1}{a} = -4$ we get :

$$\begin{aligned}\Rightarrow (-4)^3 = a^3 - \frac{1}{a^3} - 3 \times -4 \\[1em] \Rightarrow -64 = a^3 - \frac{1}{a^3} + 12 \\[1em] \Rightarrow -64 - 12 = a^3 - \frac{1}{a^3} \\[1em] \Rightarrow a^3 - \frac{1}{a^3} = -76.\end{aligned}$$

Hence, $a^3 - \frac{1}{a^3} = \pm 76$.

By formula,

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

Hence, proved that $a^3 + \frac{1}{a^3} = p(p^2 - 3)$.

Given,

⇒ a + 2b = 5

Cubing both sides we get :

⇒ (a + 2b)³ = 5³

⇒ a³ + (2b)³ + 3 × a × 2b (a + 2b) = 125

⇒ a³ + 8b³ + 6ab(a + 2b) = 125

⇒ a³ + 8b³ + 6ab × 5 = 125

⇒ a³ + 8b³ + 30ab = 125.

Hence, proved that a³ + 8b³ + 30ab = 125.

Given,

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

By formula,

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

Substituting $a + \frac{1}{a} = -\sqrt{3}$, we get :

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

Substituting $a + \frac{1}{a} = \sqrt{3}$, we get :

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

Hence, proved that $a^3 + \frac{1}{a^3} = 0.$

Given,

a + 2b + c = 0 ........(1)

By property,

If x + y + z = 0, then

⇒ x³ + y³ + z³ = 3xyz .........(2)

Comparing,

Eq. 1 with x + y + z = 0, we get :

x = a, y = 2b and z = c.

Then,

Substituting value of x, y and z in Eq. 2, we get :

⇒ a³ + (2b)³ + c³ = 3 × a × 2b × c

⇒ a³ + 8b³ + c³ = 6abc.

Hence, proved that a³ + 8b³ + c³ = 6abc.

(i) By property,

If a + b + c = 0, then

⇒ a³ + b³ + c³ = 3abc

Given,

⇒ 9³ - 5³ - 4³

⇒ 9³ + (-5)³ + (-4)³

Let a = 9, b = -5 and c = -4.

∴ a + b + c = 9 + (-5) + (-4) = 9 - 5 - 4 = 0

∴ 9³ + (-5)³ + (-4)³ = 3 × 9 × (-5) × (-4) = 540.

Hence, 9³ - 5³ - 4³ = 540.

(ii) By property,

If a + b + c = 0, then

⇒ a³ + b³ + c³ = 3abc

Given,

⇒ 38³ + (-26)³ + (-12)³

Let a = 38, b = -26 and c = -12.

∴ a + b + c = 38 + (-26) + (-12) = 38 - 26 - 12 = 0

∴ 38³ + (-26)³ + (-12)³ = 3 × 38 × (-26) × (-12) = 35,568.

Hence, 38³ + (-26)³ + (-12)³ = 35568.

(i) By formula,

$$\begin{aligned}\Rightarrow a^2 + \frac{1}{a^2} = \Big(a - \frac{1}{a}\Big)^2 + 2 \\[1em] = 3^2 + 2 \\[1em] = 9 + 2 = 11.\end{aligned}$$

Hence, $a^2 + \frac{1}{a^2} = 11$.

(ii) By formula,

$$\begin{aligned}\Rightarrow \Big(a - \frac{1}{a}\Big)^3 = a^3 - \frac{1}{a^3} - 3\Big(a -\frac{1}{a}\Big) \\[1em] \Rightarrow 3^3 = a^3 - \frac{1}{a^3} - 3 \times 3 \\[1em] \Rightarrow 27 = a^3 - \frac{1}{a^3} - 9 \\[1em] \Rightarrow a^3 - \frac{1}{a^3} = 36.\end{aligned}$$

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

(i) By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow a^2 + \frac{1}{a^2} = 4^2 + 2 \\[1em] = 16 + 2 \\[1em] = 18.\end{aligned}$$

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

(ii) By formula,

$\Rightarrow a^4 + \frac{1}{a^4} = \Big(a^2 + \frac{1}{a^2}\Big)^2 - 2$

Substituting values we get :

$$\begin{aligned}\Rightarrow a^4 + \frac{1}{a^4} = 18^2 - 2 \\[1em] = 324 - 2 \\[1em] = 322.\end{aligned}$$

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

(iii) By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow a^3 - \frac{1}{a^3} = 4^3 + 3 \times 4 \\[1em] = 64 + 12 \\[1em] = 76.\end{aligned}$$

Hence, $a^3 - \frac{1}{a^3} = 76$.

By formula,

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

Substituting values we get :

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

By formula,

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

Substituting values we get :

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

By formula,

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

Substituting values we get :

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

Hence, proved that $x^2 + \frac{1}{x^2} = x^3 + \frac{1}{x^3} = x^4 + \frac{1}{x^4}$.

Given,

2x - 3y = 10

Cubing both sides we get :

⇒ (2x - 3y)³ = 10³

⇒ (2x)³ - (3y)³ - 3 × 2x × 3y(2x - 3y) = 1000

⇒ 8x³ - 27y³ - 18xy(2x - 3y) = 1000

⇒ 8x³ - 27y³ - 18 × 16 × 10 = 1000

⇒ 8x³ - 27y³ - 2880 = 1000

⇒ 8x³ - 27y³ = 3880.

Hence, 8x³ - 27y³ = 3880.

Given,

⇒ (3x + 5y + 2z)(3x - 5y + 2z)

⇒ 9x² - 15xy + 6xz + 15xy - 25y² + 10yz + 6zx - 10yz + 4z²

⇒ 9x² - 25y² + 4z² + 12xz.

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

Given,

$$\begin{aligned}⇒ (3x - 5y - 2z)(3x - 5y + 2z) \\[1em] ⇒ 9x^2 - 15xy + 6zx - 15xy + 25y^2 - 10yz - 6zx + 10yz - 4z^2 \\[1em] ⇒ 9x^2 - 15xy + \cancel{6zx} - 15xy + 25y^2 - \cancel{10yz} - \cancel{6zx} + \cancel{10yz} - 4z^2 \\[1em] ⇒ 9x^2 - 30xy + 25y^2 - 4z^2 .\end{aligned}$$

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

Let two numbers be a and b.

Given,

Sum of numbers = 9 and Product = 20.

∴ a + b = 9 and ab = 20.

(i) Sum of squares = a² + b²

By formula,

⇒ (a + b)² = a² + b² + 2ab

Substituting values we get :

⇒ 9² = a² + b² + 2 × 20

⇒ 81 = a² + b² + 40

⇒ a² + b² = 81 - 40 = 41.

Hence, sum of squares = 41.

(ii) Sum of cubes = a³ + b³

By formula,

⇒ (a + b)³ = a³ + b³ + 3ab(a + b)

⇒ 9³ = a³ + b³ + 3 × 20 × 9

⇒ 729 = a³ + b³ + 540

⇒ a³ + b³ = 729 - 540

⇒ a³ + b³ = 189.

Hence, a³ + b³ = 189.

Given,

Difference of numbers = 5 and product = 24.

Let the two numbers be x and y.

∴ x - y = 5 and xy = 24.

(i) By formula,

(x + y)² = (x - y)² + 4xy

Substituting values we get :

⇒ (x + y)² = 5² + 4 × 24

⇒ (x + y)² = 25 + 96

⇒ (x + y)² = 121

⇒ (x + y) = $\sqrt{121} = \pm 11$.

Since, numbers are positive so sum cannot be negative.

Hence, sum of these numbers = 11.

(ii) By formula,

⇒ x³ - y³ = (x - y)³ + 3xy(x - y)

⇒ x³ - y³ = 5³ + 3 × 24 × 5

⇒ x³ - y³ = 125 + 360

⇒ x³ - y³ = 485.

Hence, difference of cubes of numbers = 485.

(iii) By formula,

⇒ x³ + y³ = (x + y)³ - 3xy(x + y)

Substituting x + y = 11, we get :

⇒ x³ + y³ = 11³ - 3 × 24 × 11

⇒ x³ + y³ = 1331 - 792

⇒ x³ + y³ = 539.

Hence, sum of cubes of numbers = 539.

⇒ (2x + y)² = (2x)² + y² + 2 × 2x × y

⇒ (2x + y)² = 4x² + y² + 4xy

Substituting values we get,

⇒ (2x + y)² = a + 4b

⇒ (2x + y) = $\pm \sqrt{a + 4b}$.

Hence, (2x + y) = $\pm \sqrt{a + 4b}$.

Expansion of (x - a)(x + b) = x² - (a - b)x - ab

∴ Expansion of (x - 3y)(x + 5y) = x² - (3y - 5y)x - 3y × 5y

= x² - (-2y)x - 15y²

= x² + 2xy - 15y².

Hence, Option 2 is the correct option.

Given,

⇒ x² - (a + b)x + ab

⇒ x² - ax - bx + ab

⇒ x(x - a) - b(x - a)

⇒ (x - a)(x - b).

Hence, Option 1 is the correct option.

By formula,

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

Substituting values we get :

⇒ 4² = 14 + 2(ab - bc - ca)

⇒ 16 = 14 + 2(ab - bc - ca)

⇒ 2(ab - bc - ca) = 16 - 14

⇒ 2(ab - bc - ca) = 2

⇒ ab - bc - ca = 1

Hence, Option 2 is the correct option.

Given,

⇒ $\Big(a - \frac{1}{2a}\Big)^2$

Expanding,

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

Hence, Option 3 is the correct option.

Given,

⇒ (x + 8)(x + 10)

Expanding,

⇒ x² + 10x + 8x + 80

⇒ x² + 18x + 80.

Hence, expansion of (x + 8)(x + 10) = x² + 18x + 80.

Given,

⇒ (x + 8)(x - 10)

Expanding,

⇒ x² - 10x + 8x - 80

⇒ x² - 2x - 80.

Hence, expansion of (x + 8)(x - 10) = x² - 2x - 80.

Given,

⇒ (x - 8)(x + 10)

Expanding,

⇒ x² + 10x - 8x - 80

⇒ x² + 2x - 80.

Hence, expansion of (x - 8)(x + 10) = x² + 2x - 80.

Given,

⇒ (x - 8)(x - 10)

Expanding,

⇒ x² - 10x - 8x + 80

⇒ x² - 18x + 80.

Hence, expansion of (x - 8)(x - 10) = x² - 18x + 80.

Given,

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

Expanding,

$$\begin{aligned}\Rightarrow 2x \times 3x + 2x \times \frac{2}{x} - \frac{1}{x} \times 3x - \frac{1}{x} \times \frac{2}{x} \\[1em] \Rightarrow 6x^2 + 4 - 3 - \frac{2}{x^2} \\[1em] \Rightarrow 6x^2 + 1 - \frac{2}{x^2}.\end{aligned}$$

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

Given,

$\Big(3a + \frac{2}{b}\Big)\Big(2a - \frac{3}{b}\Big)$

Expanding,

$$\begin{aligned}\Rightarrow 3a \times 2a + 3a \times -\frac{3}{b} + \frac{2}{b} \times 2a + \frac{2}{b} \times -\frac{3}{b} \\[1em] \Rightarrow 6a^2 - \frac{9a}{b} + \frac{4a}{b} - \frac{6}{b^2} \\[1em] \Rightarrow 6a^2 - \frac{5a}{b} - \frac{6}{b^2}.\end{aligned}$$

Hence, $\Big(3a + \frac{2}{b}\Big)\Big(2a - \frac{3}{b}\Big) = 6a^2 - \frac{5a}{b} - \frac{6}{b^2}.$

Given,

⇒ (x + y - z)²

Expanding,

⇒ x² + y² + z² + 2xy - 2yz - 2zx

⇒ x² + y² + z² + 2(xy - yz - zx).

Hence, (x + y - z)² = x² + y² + z² + 2(xy - yz - zx).

Given,

⇒ (x - 2y + 2)²

Expanding,

⇒ (x - 2y + 2)(x - 2y + 2)

⇒ x² - 2xy + 2x - 2xy + 4y² - 4y + 2x - 4y + 4

⇒ x² + 4y² + 4 - 4xy - 8y + 4x.

Hence, (x - 2y + 2)² = x² + 4y² + 4 - 4xy - 8y + 4x.

Given,

⇒ (5a - 3b + c)²

Expanding,

⇒ (5a - 3b + c)(5a - 3b + c)

⇒ (5a)² - 15ab + 5ac - 15ab + 9b² - 3bc + 5ac - 3bc + c²

⇒ 25a² + 9b² + c² - 30ab - 6bc + 10ac.

Hence, (5a - 3b + c)² = 25a² + 9b² + c² - 30ab - 6bc + 10ac.

Given,

⇒ (5x - 3y - 2)²

Expanding,

⇒ (5x - 3y - 2)(5x - 3y - 2)

⇒ 25x² - 15xy - 10x - 15xy + 9y² + 6y - 10x + 6y + 4

⇒ 25x² + 9y² - 30xy - 20x + 12y + 4.

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

Given,

⇒ $\Big(x - \frac{1}{x} + 5\Big)^2$

Expanding,

$$\begin{aligned}\Rightarrow \Big(x - \frac{1}{x} + 5\Big)\Big(x - \frac{1}{x} + 5\Big) \\[1em] \Rightarrow x^2 - 1 + 5x - 1 + \frac{1}{x^2} - \frac{5}{x} + 5x - \frac{5}{x} + 25 \\[1em] \Rightarrow x^2 + \frac{1}{x^2} - \frac{10}{x} + 23 + 10x.\end{aligned}$$

Hence, $\Big(x - \frac{1}{x} + 5\Big)^2 = x^2 + \frac{1}{x^2} - \frac{10}{x} + 23 + 10x.$

By formula,

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

Substituting values we get :

⇒ 12² = 50 + 2(ab + bc + ca)

⇒ 144 = 50 + 2(ab + bc + ca)

⇒ 2(ab + bc + ca) = 144 - 50

⇒ 2(ab + bc + ca) = 94

⇒ ab + bc + ca = 47.

Hence, ab + bc + ca = 47.

By formula,

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

Substituting values we get :

⇒ (a + b + c)² = 35 + 2 × 23

⇒ (a + b + c)² = 35 + 46

⇒ (a + b + c)² = 81

⇒ (a + b + c) = $\sqrt{81} = \pm 9$.

Hence, (a + b + c) = $\pm 9$.

By formula,

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

Substituting values we get :

⇒ p² = a² + b² + c² + 2q

⇒ a² + b² + c² = p² - 2q.

Hence, a² + b² + c² = p² - 2q.

By formula,

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

Substituting values we get :

⇒ (a + b + c)² = 50 + 2 × 47

⇒ (a + b + c)² = 50 + 94

⇒ (a + b + c)² = 144

⇒ (a + b + c) = $\sqrt{144} = \pm 12$.

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

By formula,

(x + y - z)² = x² + y² + z² + 2(xy - yz - zx)

Substituting values we get :

⇒ 4² = 30 + 2(xy - yz - zx)

⇒ 16 = 30 + 2(xy - yz - zx)

⇒ 2(xy - yz - zx) = 16 - 30

⇒ 2(xy - yz - zx) = -14

⇒ xy - yz - zx = -7.

Hence, xy - yz - zx = -7.

Given, the longest road that can be placed in a rectangular box, d = 20 cm

The longest rod which can be kept inside a rectangular box will be equal to the diagonal of the box.

By formula,

⇒ Diagonal² = l² + b² + h²

⇒ 20² = l² + b² + h²

⇒ 400 = l² + b² + h² ..................(1)

Given,

The sum of its length breadth and height equals to 30 cm.

⇒ l + b + h = 30 cm.

Squaring both sides, we get :

⇒ (l + b + h)² = 30²

⇒ l² + b² + h² + 2(lb + bh + hl) = 900

From equation (1), we get

⇒ 400 + 2(lb + bh + hl) = 900

⇒ 2(lb + bh + hl) = 900 - 400

⇒ 2(lb + bh + hl) = 500

Hence, the total surface area of the box = 500 cm².

Solving,

$$\begin{aligned}\Rightarrow \Big(a + \frac{1}{a}\Big)^2 - \Big(a - \frac{1}{a}\Big)^2 \\[1em] \Rightarrow (2.5)^2 - (1.5)^2 \\[1em] \Rightarrow 6.25 - 2.25 \\[1em] \Rightarrow 4.\end{aligned}$$

Hence, Option 1 is the correct option.

Given,

⇒ (3x + 2)(x - 4)

⇒ 3x² - 12x + 2x - 8

⇒ 3x² - 10x - 8.

∴ Coefficient of x is -10.

Hence, Option 3 is the correct option.

Given,

⇒ (x + y - z)(x - y + z)

Expanding,

⇒ x² - xy + xz + xy - y² + yz - zx + yz - z²

⇒ x² - y² - z² + 2yz.

Hence, Option 2 is the correct option.

Solving,

⇒ (3x - 4y)² = 9x² + axy + 16y²

⇒ 9x² + 16y² - 24xy = 9x² + axy + 16y²

From above equation,

axy = -24xy

a = $-\frac{24xy}{xy}$ = -24.

Hence, Option 1 is the correct option.

Given,

x + 2y + 3z = 0

∴ x + 2y = -3z, 2y + 3z = -x and 3z + x = -2y

Solving,

$$\begin{aligned}\Rightarrow \frac{(x + 2y)^2}{xy} + \frac{(2y + 3z)^2}{yz} + \frac{(3z + x)^2}{zx} \\[1em] \Rightarrow \frac{(-3z)^2}{xy} + \frac{(-x)^2}{yz} + \frac{(-2y)^2}{zx}\\[1em] \Rightarrow \frac{9z^2}{xy} + \frac{x^2}{yz} + \frac{4y^2}{zx} \\[1em] \Rightarrow \frac{9z^3 + x^3 + 4y^3}{xyz} \\[1em] \Rightarrow \frac{x^3 + 4y^3 + 9z^3}{xyz} \\[1em] \Rightarrow \frac{18xyz}{xyz} \\[1em] \Rightarrow 18.\end{aligned}$$

Hence, $\frac{(x + 2y)^2}{xy} + \frac{(2y + 3z)^2}{yz} + \frac{(3z + x)^2}{zx}$ = 18.

(i) By formula,

$\Rightarrow \Big(a + \frac{1}{a}\Big)^2 - \Big(a - \frac{1}{a}\Big)^2 = 4$

Substituting values we get :

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

Hence, $a - \frac{1}{a} = \pm \sqrt{m^2 - 4}.$

(ii) By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow \Rightarrow a^2 - \frac{1}{a^2} = \pm \sqrt{m^2 - 4} \times m \\[1em] = \pm m\sqrt{m^2 - 4}.\end{aligned}$$

Hence, $a^2 - \frac{1}{a^2} = \pm m\sqrt{m^2 - 4}.$

Given,

⇒ (2x² - 8)(x - 4)²

Expanding,

⇒ (2x² - 8)(x² + 16 - 8x)

⇒ 2x⁴ + 32x² - 16x³ - 8x² - 128 + 64x

⇒ 2x⁴ - 16x³ + 24x² + 64x - 128.

(i) Hence, coefficient of x³ is -16.

(ii) Hence, coefficient of x² is 24.

(iii) Hence, constant term = -128.

Given,

$$\begin{aligned}\Rightarrow x^2 + \frac{1}{9x^2} = \frac{25}{36} \\[1em] \Rightarrow \Big(x + \frac{1}{3x}\Big)^2 - \frac{2}{3} = \frac{25}{36} \\[1em] \Rightarrow \Big(x + \frac{1}{3x}\Big)^2 = \frac{25}{36} + \frac{2}{3} \\[1em] \Rightarrow \Big(x + \frac{1}{3x}\Big)^2 = \frac{25 + 24}{36} \\[1em] \Rightarrow \Big(x + \frac{1}{3x}\Big)^2 = \frac{49}{36}\\[1em] \Rightarrow x + \frac{1}{3x} = \sqrt{\frac{49}{36}} \\[1em] \Rightarrow x + \frac{1}{3x} = \pm \frac{7}{6}.\end{aligned}$$

Since, x is > 0,

∴ $x + \frac{1}{3x} = \frac{7}{6}$

By formula,

$\Rightarrow \Big(x^3 + \frac{1}{27x^3}\Big) = \Big(x + \frac{1}{3x}\Big)^3 - \Big(x + \frac{1}{3x}\Big)$ .......(1)

Substituting $x + \frac{1}{3x} = \frac{7}{6}$ in equation (1), we get :

$$\begin{aligned}\Rightarrow \Big(x^3 + \frac{1}{27x^3}\Big) = \Big(\frac{7}{6}\Big)^3 - \Big(\frac{7}{6}\Big) \\[1em] = \frac{343}{216} - \frac{7}{6} \\[1em] = \frac{343 - 252}{216} \\[1em] = \frac{91}{216}.\end{aligned}$$

Hence, $x^3 + \frac{1}{27x^3} = \frac{91}{216}$.

(i) Given,

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

By formula,

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{x}\Big)^2 - \Big(x - \frac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(\frac{5}{2}\Big)^2 - \Big(x - \frac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big)^2 = \frac{25}{4} - 4\\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big)^2 = \frac{25 - 16}{4} \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big)^2 = \frac{9}{4} \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big) = \sqrt{\frac{9}{4}} \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big) = \pm \frac{3}{2}.\end{aligned}$$

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

(ii) By formula,

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

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

$$\begin{aligned}\Rightarrow \Big(x^3 - \frac{1}{x^3}\Big) = \Big(\frac{3}{2}\Big)^3 + 3 \times \frac{3}{2} \\[1em] = \frac{27}{8} + \frac{9}{2} \\[1em] = \frac{27 + 36}{8} \\[1em] = \frac{63}{8}.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \Big(x^3 - \frac{1}{x^3}\Big) = \Big(-\frac{3}{2}\Big)^3 + 3 \times -\frac{3}{2} \\[1em] = -\frac{27}{8} - \frac{9}{2} \\[1em] = \frac{-27 - 36}{8} \\[1em] = \frac{-63}{8}.\end{aligned}$$

Hence, $\Big(x^3 - \frac{1}{x^3}\Big) = \pm \frac{63}{8}$.

(i) Expanding,

⇒ 3(a + b)² + 5(a - b)²

⇒ 3(a² + b² + 2ab) + 5(a² + b² - 2ab)

⇒ 3(34 + 2 × 12) + 5(34 - 2 × 12)

⇒ 3(34 + 24) + 5(34 - 24)

⇒ 3 × 58 + 5 × 10

⇒ 174 + 50

⇒ 224.

Hence, 3(a + b)² + 5(a - b)² = 224.

(ii) Expanding,

⇒ 7(a - b)² - 2(a + b)²

⇒ 7(a² + b² - 2ab) - 2(a² + b² + 2ab)

⇒ 7(34 - 2 × 12) - 2(34 + 2 × 12)

⇒ 7(34 - 24) - 2(34 + 24)

⇒ 7 × 10 - 2 × 58

⇒ 70 - 116

⇒ -46.

Hence, 7(a - b)² - 2(a + b)² = -46.

Given,

$\Rightarrow 3x - \frac{4}{x} = 4$

Cubing both sides we get :

$$\begin{aligned}\Rightarrow \Big(3x - \frac{4}{x}\Big)^3 = 4^3 \\[1em] \Rightarrow (3x)^3 - \Big(\frac{4}{x}\Big)^3 - 3 \times 3x \times \frac{4}{x} \times \Big(3x - \frac{4}{x}\Big) = 64 \\[1em] \Rightarrow (3x)^3 - \Big(\frac{4}{x}\Big)^3 - 36 \times 4 = 64 \\[1em] \Rightarrow 27x^3 - \frac{64}{x^3} - 144 = 64 \\[1em] \Rightarrow 27x^3 - \frac{64}{x^3} = 64 + 144 \\[1em] \Rightarrow 27x^3 - \frac{64}{x^3} = 208.\end{aligned}$$

Hence, $27x^3 - \frac{64}{x^3} = 208.$

By formula,

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

By formula,

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

Substituting $x - \frac{1}{x} = \sqrt{5}$ in above equation, we get :

$$\begin{aligned}\Rightarrow (\sqrt{5})^3 = x^3 - \frac{1}{x^3} - 3 \times \sqrt{5} \\[1em] \Rightarrow 5\sqrt{5} = x^3 - \frac{1}{x^3} - 3\sqrt{5} \\[1em] \Rightarrow x^3 - \frac{1}{x^3} = 8\sqrt{5}.\end{aligned}$$

Substituting $x - \frac{1}{x} = -\sqrt{5}$ in above equation, we get :

$$\begin{aligned}\Rightarrow (-\sqrt{5})^3 = x^3 - \frac{1}{x^3} - 3 \times -\sqrt{5} \\[1em] \Rightarrow -5\sqrt{5} = x^3 - \frac{1}{x^3} + 3\sqrt{5} \\[1em] \Rightarrow x^3 - \frac{1}{x^3} = -8\sqrt{5}.\end{aligned}$$

Solving given equation,

$$\begin{aligned}\Rightarrow 7x^3 + 8x - \frac{7}{x^3} - \frac{8}{x} \\[1em] \Rightarrow 7x^3 - \frac{7}{x^3} + 8x - \frac{8}{x} \\[1em] \Rightarrow 7\Big(x^3 - \frac{1}{x^3}\Big) + 8\Big(x - \frac{1}{x}\Big)\end{aligned}$$

Substituting $x^3 - \frac{1}{x^3} = 8\sqrt{5}\text{ and } x - \frac{1}{x} = \sqrt{5}$, we get :

$$\begin{aligned}\Rightarrow 7 \times 8\sqrt{5} + 8 \times \sqrt{5} \\[1em] \Rightarrow 56\sqrt{5} + 8\sqrt{5} \\[1em] \Rightarrow 64\sqrt{5}.\end{aligned}$$

Substituting $x^3 - \frac{1}{x^3} = -8\sqrt{5}\text{ and } x - \frac{1}{x} = -\sqrt{5}$, we get :

$$\begin{aligned}\Rightarrow 7 \times -8\sqrt{5} + 8 \times -\sqrt{5} \\[1em] \Rightarrow -56\sqrt{5} - 8\sqrt{5} \\[1em] \Rightarrow -64\sqrt{5}.\end{aligned}$$

Hence, 7x³ + 8x - $\frac{7}{x^3} - \frac{8}{x} = \pm 64\sqrt{5}$.

Given,

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

By formula,

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

By formula,

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

Hence, $x^2 - \frac{1}{x^2} = \pm 5\sqrt{29}$.

Given,

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

By formula,

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

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

By property,

If x + y + z = 0, then :

x³ + y³ + z³ = 3xyz ........(1)

Comparing,

3a + 5b + 4c = 0 with x + y + z = 0, we get :

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

Substituting values in equation (1), we get :

⇒ (3a)³ + (5b)³ + (4c)³ = 3 × 3a × 5b × 4c

⇒ 27a³ + 125b³ + 64c³ = 180abc.

Hence, proved that 27a³ + 125b³ + 64c³ = 180abc.

Let two numbers be x and y.

Given,

The sum of two numbers is 7 and the sum of their cubes is 133.

x + y = 7 and x³ + y³ = 133

By formula,

⇒ x³ + y³ = (x + y)³ - 3xy(x + y)

⇒ 133 = 7³ - 3xy × 7

⇒ 133 = 343 - 21xy

⇒ 21xy = 343 - 133

⇒ 21xy = 210

⇒ xy = 10.

By formula,

⇒ (x + y)² = x² + y² + 2xy

⇒ 7² = x² + y² + 2 × 10

⇒ 49 = x² + y² + 20

⇒ x² + y² = 49 - 20

⇒ x² + y² = 29.

Hence, sum of squares of numbers = 29.

(i) Given,

⇒ 4x² + ax + 9 = (2x + 3)²

⇒ 4x² + ax + 9 = (2x)² + 3² + 2 × 2x × 3

⇒ 4x² + ax + 9 = 4x² + 9 + 12x

⇒ ax = 12x

⇒ a = 12.

Hence, a = 12.

(ii) Given,

⇒ 4x² + ax + 9 = (2x - 3)²

⇒ 4x² + ax + 9 = (2x)² + 3² - 2 × 2x × 3

⇒ 4x² + ax + 9 = 4x² + 9 - 12x

⇒ ax = -12x

⇒ a = -12.

Hence, a = -12.

(iii) Given,

⇒ 9x² + (7a - 5)x + 25 = (3x + 5)²

⇒ 9x² + (7a - 5)x + 25 = (3x)² + 5² + 2 × 3x × 5

⇒ 9x² + (7a - 5)x + 25 = 9x² + 30x + 25

⇒ 7a - 5 = 30

⇒ 7a = 35

⇒ a = $\frac{35}{7}$ = 5.

Hence, a = 5.

(i) Given,

$$\begin{aligned}\Rightarrow \frac{x^2 + 1}{x} = 3\frac{1}{3} \\[1em] \Rightarrow x + \frac{1}{x} = \frac{10}{3}.\end{aligned}$$

By formula,

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{x}\Big)^2 - \Big(x - \frac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(\frac{10}{3}\Big)^2 - \Big(x - \frac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \frac{100}{9} - \Big(x - \frac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big)^2 = \frac{100}{9} - 4 \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big)^2 = \frac{100 - 36}{9} \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big)^2 = \frac{64}{9} \\[1em] \Rightarrow x - \frac{1}{x} = \sqrt{\frac{64}{9}} \\[1em] \Rightarrow x - \frac{1}{x} = \frac{8}{3} = 2\frac{2}{3}.\end{aligned}$$

Hence, $x - \frac{1}{x} = 2\frac{2}{3}$.

(ii) By formula,

$$\begin{aligned}\Rightarrow x^3 - \frac{1}{x^3} = \Big(x - \frac{1}{x}\Big)^3 + 3\Big(x - \frac{1}{x}\Big) \\[1em] = \Big( \frac{8}{3}\Big)^3 + 3 \times \frac{8}{3} \\[1em] = \frac{512}{27} + \frac{24}{3} \\[1em] = \frac{512 + 216}{27} \\[1em] = \frac{728}{27} \\[1em] = 26\frac{26}{27}.\end{aligned}$$

Hence, $x^3 - \frac{1}{x^3} = 26\frac{26}{27}$.

Let two positive numbers be x and y with x > y.

Given,

The difference between two positive numbers is 4 and the difference between their cubes is 316.

x - y = 4 and x³ - y³ = 316.

(i) Given,

⇒ x - y = 4

Cubing both sides we get :

⇒ (x - y)³ = 4³

⇒ x³ - y³ - 3xy(x - y) = 64

⇒ 316 - 3xy × 4 = 64

⇒ 316 - 12xy = 64

⇒ 12xy = 316 - 64

⇒ 12xy = 252

⇒ xy = $\frac{252}{12}$ = 21.

Hence, product of numbers = 21.

(ii) Given,

x - y = 4

Squaring both sides we get :

⇒ (x - y)² = 4²

⇒ x² + y² - 2xy = 16

⇒ x² + y² - 2 × 21 = 16

⇒ x² + y² - 42 = 16

⇒ x² + y² = 16 + 42

⇒ x² + y² = 58.

Hence, sum of squares = 58.

Given, x² + $\frac{1}{\text{x}^2}$ = 3

As we know

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

Hence, option 3 is the correct option.

Given, a³ + b³ + c³ = 3abc

By formula,

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

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

⇒ 0 = (a + b + c)(a² + b² + c² - ab - bc - ca)

⇒ (a + b + c) = 0 or (a² + b² + c² - ab - bc - ca) = 0

Hence, option 4 is the correct option.

Given, $\frac{196 \times 196 \times 196 + 204 \times 204 \times 204}{(196)^2 + (204)^2 - 196 \times 204}$

Using the formula, (a³ + b³) = (a + b)(a² - ab + b²)

$$\begin{aligned}\Rightarrow \frac{196^3 + 204^3}{(196)^2 + (204)^2 - 196 \times 204}\\[1em] = \frac{\Big(196 + 204\Big)\Big((196)^2 + (204)^2 - 196 \times 204\Big)}{(196)^2 + (204)^2 - 196 \times 204}\\[1em] = (196 + 204)\\[1em] = 400.\end{aligned}$$

Hence, option 1 is the correct option.

Given, a = x + $\frac{1}{\text{x}}$ and b = x - $\frac{1}{\text{x}}$

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

Hence, option 3 is the correct option.

Given, x² + $\frac{1}{\text{x}^2}$ = 2

As we know,

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

Now, we know that $(x^2 - \frac{1}{x^2}) = (x + \frac{1}{x})(x - \frac{1}{x})$

= $\pm$ 2 x 0

= 0.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Given, x² - 5x - 1 = 0

⇒ x² - 1 = 5x

⇒ $\frac{x^2 - 1}{x}$ = 5

⇒ $\frac{x^2}{x} - \frac{1}{x}$ = 5

⇒ x - $\frac{1}{x}$ = 5

So, x - $\frac{1}{x}$ = 5 is true when x ≠ 0

∴ A is false, but R is true.

Hence, option 2 is the correct option.

We know,

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

So, 3abc − a³ − b³ − c³ = −(a³ + b³+ c³ −3abc)

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

Given, $\frac{ab + bc + ca - a^2 - b^2 - c^2}{3abc - a^3 - b^3 - c^3}$

= $\frac{-(a^2 + b^2 + c^2 − ab − bc − ca )}{−(a + b + c)(a^2 + b^2 + c^2 − ab − bc − ca)}$

= $\frac{(a^2 + b^2 + c^2 − ab − bc − ca )}{(a + b + c)(a^2 + b^2 + c^2 − ab − bc − ca)}$

= $\frac{1}{a + b + c}$

So, assertion (A) is true.

By formula,

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

Given expression, $\frac{ab + bc + ca - a^2 - b^2 - c^2}{3abc - a^3 - b^3 - c^3}$

= $\frac{a^2 + b^2 + c^2 - ab - bc - ca}{a^3 + b^3 + c^3 - 3abc}$

= $\frac{a^2 + b^2 + c^2 - ab - bc - ca}{(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)}$

So, Reason (R) is true.

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

By formula,

⇒ (x + y)² = x² + y² + 2xy ....(1)

⇒ (x - y)² = x² + y² - 2xy .....(2)

Adding equation (1) and (2), we get :

⇒ (x + y)² + (x - y)² = x² + y² + 2xy + x² + y² - 2xy

⇒ (x + y)² + (x - y)² = 2(x² + y²)

So, reason (R) is true.

Given,

x + y = 6 and x - y = 2

By formula,

⇒ 2(x² + y²) = (x + y)² + (x - y)²

⇒ 2(x² + y²) = 6² + 2²

⇒ 2(x² + y²) = 36 + 4

⇒ 2(x² + y²) = 40

⇒ x² + y² = $\frac{40}{2}$

⇒ x² + y² = 20.

So, assertion (A) is false.

∴ A is false, but R is true.

Hence, option 2 is the correct option.

By property,

If a + b + c = 0, then

⇒ a³ + b³ + c³ = 3abc

Given,

⇒ 5³ - 3³ - 2³

⇒ 5³ + (-3)³ + (-2)³

Let a = 5, b = -3 and c = -2.

∴ a + b + c = 5 + (-3) + (-2) = 5 - 3 - 2 = 0

⇒ a³ + b³ + c³ = 3abc

⇒ 5³ + (-3)³ + (-2)³ = 3 × 5 × (-3) × (-2)

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

Given,

⇒ (x + 6)(x + 4)(x - 2)

⇒ (x + 6)(x² - 2x + 4x - 8)

⇒ (x + 6)(x² + 2x - 8)

⇒ x³ + 2x² - 8x + 6x² + 12x - 48

⇒ x³ + 8x² + 4x - 48.

Hence, (x + 6)(x + 4)(x - 2) = x³ + 8x² + 4x - 48.

Given,

⇒ (x - 6)(x - 4)(x + 2)

⇒ (x - 6)(x² + 2x - 4x - 8)

⇒ (x - 6)(x² - 2x - 8)

⇒ x³ - 2x² - 8x - 6x² + 12x + 48

⇒ x³ - 8x² + 4x + 48.

Hence, (x - 6)(x - 4)(x + 2) = x³ - 8x² + 4x + 48.

Given,

⇒ (x - 6)(x - 4)(x - 2)

⇒ (x - 6)(x² - 2x - 4x + 8)

⇒ (x - 6)(x² - 6x + 8)

⇒ x³ - 6x² + 8x - 6x² + 36x - 48

⇒ x³ - 12x² + 44x - 48.

Hence, (x - 6)(x - 4)(x - 2) = x³ - 12x² + 44x - 48.

Given,

⇒ (x + 6)(x - 4)(x - 2)

⇒ (x + 6)(x² - 2x - 4x + 8)

⇒ (x + 6)(x² - 6x + 8)

⇒ x³ - 6x² + 8x + 6x² - 36x + 48

⇒ x³ - 28x + 48.

Hence, (x + 6)(x - 4)(x - 2) = x³ - 28x + 48.

(i) Given,

⇒ (2x + 3y)(4x² - 6xy + 9y²)

⇒ (2x + 3y)[(2x)² - 2x × 3y + (3y)²]

Comparing above equation with (a $\pm$ b)(a² $\pm$ ab $\pm$ b²) = a³ $\pm$ b³, we get :

a = 2x and b = 3y

∴ (2x + 3y)[(2x)² - 2x × 3y + (3y)²] = (2x)³ + (3y)³

= 8x³ + 27y³.

Hence, (2x + 3y)(4x² - 6xy + 9y²) = 8x³ + 27y³.

(ii) Given,

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

Using identity given in question :

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

Hence, $\Big(\frac{a}{3} - 3b\Big)\Big(\frac{a^2}{9} + ab + 9b^2\Big) = \frac{a^3}{27} - 27b^3$.

(i) Given,

⇒ (104)³

⇒ (100 + 4)³

Using identity :

(a + b)³ = a³ + b³ + 3ab(a + b)

⇒ (100 + 4)³ = (100)³ + 4³ + 3 × 100 × 4 × (100 + 4)

⇒ (100 + 4)³ = 1000000 + 64 + 124800

⇒ 1124864.

Hence, (104)³ = 1124864.

(ii) Given,

⇒ (97)³

⇒ (100 - 3)³

⇒ (100)³ - 3³ - 3 × 100 × 3 × (100 - 3)

⇒ 1000000 - 27 - 900 × 97

⇒ 1000000 - 27 - 87300

⇒ 912673.

Hence, (97)³ = 912673.

If a + b + c = 0; we have :

a³ + b³ + c³ = 3abc.

Since, (x² - y²) + (y² - z²) + (z² - x²) = 0.

∴ $(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3 = 3(x^2 - y^2)(y^2 - z^2)(z^2 - x^2)$ .........(1)

Also,

(x - y) + (y - z) + (z - x) = 0

∴ $(x - y)^3 + (y- z)^3 + (z - x)^3 = 3(x - y)(y - z)(z - x)$ ..............(2)

Dividing equation (1) by (2), we get :

$$\begin{aligned}\Rightarrow \frac{(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3}{(x - y)^3 + (y- z)^3 + (z - x)^3} = \frac{3(x^2 - y^2)(y^2 - z^2)(z^2 - x^2)}{3(x - y)(y - z)(z - x)} \\[1em] = \frac{3(x - y)(x + y)(y - z)(y + z)(z - x)(z + x)}{3(x - y)(y - z)(z - x)} \\[1em] = (x + y)(y + z)(z + x).\end{aligned}$$

Hence, $\frac{(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3}{(x - y)^3 + (y- z)^3 + (z - x)^3}$ = (x + y)(y + z)(z + x).

Given,

$$\begin{aligned}\Rightarrow \frac{0.8 \times 0.8 \times 0.8 + 0.5 \times 0.5 \times 0.5}{0.8 \times 0.8 - 0.8 \times 0.5 + 0.5 \times 0.5} \\[1em] \Rightarrow \frac{(0.8)^3 + (0.5)^3}{(0.8)^2 + (0.5)^2 - 0.8 \times 0.5}\end{aligned}$$

Using the formula, (a³ + b³) = (a + b)(a² - ab + b²)

$$\begin{aligned}= \frac{(0.8 + 0.5)[(0.8)^2 + (0.5)^2 - 0.8 \times 0.5]}{(0.8)^2 + (0.5)^2 - 0.8 \times 0.5}\\[1em] = (0.8 + 0.5)\\[1em] = 1.3\end{aligned}$$

Hence, $\frac{0.8 \times 0.8 \times 0.8 + 0.5 \times 0.5 \times 0.5}{0.8 \times 0.8 - 0.8 \times 0.5 + 0.5 \times 0.5}$ = 1.3.

By property,

If x + y + z = 0, then :

x³ + y³ + z³ = 3xyz.

Given,

⇒ a - 2b + 3c = 0

⇒ a + (-2b) + 3c = 0

∴ a³ + (-2b)³ + (3c)³ = 3 × a × (-2b) × 3c

⇒ a³ - 8b³ + 27c³ = -18abc.

Hence, a³ - 8b³ + 27c³ = -18abc.

Given,

⇒ x + 5y = 10

Cubing both sides we get :

⇒ (x + 5y)³ = 10³

⇒ x³ + (5y)³ + 3 × x × 5y × (x + 5y) = 1000

⇒ x³ + 125y³ + 15xy × 10 = 1000

⇒ x³ + 125y³ + 150xy - 1000 = 0.

Hence, x³ + 125y³ + 150xy - 1000 = 0.

By formula,

⇒ (a + b)² = a² + b² + 2ab

⇒ 11² = 65 + 2ab

⇒ 121 = 65 + 2ab

⇒ 2ab = 56

⇒ ab = $\frac{56}{2}$ = 28.

By formula,

⇒ (a + b)³ = a³ + b³ + 3ab(a + b)

⇒ 11³ = a³ + b³ + 3 × 28 × 11

⇒ 1331 = a³ + b³ + 924

⇒ a³ + b³ = 1331 - 924 = 407.

Hence, a³ + b³ = 407.

Given,

⇒ x² + y² + z² - xy - yz - zx

Multiplying the above equation by 2,

⇒ 2(x² + y² + z² - xy - yz - zx )

⇒ 2x² + 2y² + 2z² - 2xy - 2yz - 2zx

⇒ x² + x² + y² + y² + z² + z² - 2xy - 2yz - 2zx

⇒ x² + y² - 2xy + y² + z² - 2yz + z² + x² - 2zx

⇒ (x - y)² + (y - z)² + (z - x)²

From above equation we can see that for distinct value of x, y and z given equation is always positive.

Given,

x = $\sqrt[3]{65}$

y = $\sqrt[3]{64}$

By formula,

x³ - y³ = (x - y) (x² + xy + y²)

x³ - y³ = $(\sqrt[3]{65})^3$ - $(\sqrt[3]{64})^3$ = 65 - 64 = 1

∴ 1 = (x - y) (x² + xy + y²)......(1)

Solving,

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

Hence, $(x - y) - \frac{1}{x^2 + xy + y^2}$ = 0.