Algebraic Identities

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (x + 5)(x + 7) \\[1em] \Rightarrow x^2 + (5 + 7)x + 5 \times 7 \\[1em] \Rightarrow x^2 + 12x + 35\end{aligned}$$

Hence, $\bm{(x + 5)(x + 7) = x^2 + 12x + 35}$.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (b + 2)(b + 9) \\[1em] \Rightarrow b^2 + (2 + 9)b + 2 \times 9 \\[1em] \Rightarrow b^2 + 11b + 18\end{aligned}$$

Hence, $\bm{(b + 2)(b + 9) = b^2 + 11b + 18}$.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (c + 2)\Big(c + \frac{3}{5}\Big) \\[1em] \Rightarrow c^2 + \Big(2 + \frac{3}{5}\Big)c + 2 \times \frac{3}{5} \\[1em] \Rightarrow c^2 + \frac{10 + 3}{5}c + \frac{6}{5} \\[1em] \Rightarrow c^2 + \frac{13}{5}c + \frac{6}{5}\end{aligned}$$

Hence, $\bm{(c + 2)\Big(c + \frac{3}{5}\Big) = c^2 + \frac{13}{5}c + \frac{6}{5}}$.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

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

Hence, $\bm{\Big(t + \frac{4}{3}\Big)\Big(t + \frac{1}{3}\Big) = t^2 + \frac{5}{3}t + \frac{4}{9}}$.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (y + 8)(y - 4) \\[1em] \Rightarrow (y + 8)[y + (-4)] \\[1em] \Rightarrow y^2 + [8 + (-4)]y + 8 \times (-4) \\[1em] \Rightarrow y^2 + 4y - 32\end{aligned}$$

Hence, $\bm{(y + 8)(y - 4) = y^2 + 4y - 32}$.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (z + 6)(z - 11) \\[1em] \Rightarrow (z + 6)[z + (-11)] \\[1em] \Rightarrow z^2 + [6 + (-11)]z + 6 \times (-11) \\[1em] \Rightarrow z^2 - 5z - 66\end{aligned}$$

Hence, $\bm{(z + 6)(z - 11) = z^2 - 5z - 66}$.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (c - 5)(c + 1) \\[1em] \Rightarrow [c + (-5)](c + 1) \\[1em] \Rightarrow c^2 + [(-5) + 1]c + (-5) \times 1 \\[1em] \Rightarrow c^2 - 4c - 5\end{aligned}$$

Hence, $\bm{(c - 5)(c + 1) = c^2 - 4c - 5}$.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (b - 13)(b + 10) \\[1em] \Rightarrow [b + (-13)](b + 10) \\[1em] \Rightarrow b^2 + [(-13) + 10]b + (-13) \times 10 \\[1em] \Rightarrow b^2 - 3b - 130\end{aligned}$$

Hence, $\bm{(b - 13)(b + 10) = b^2 - 3b - 130}$.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (x - 3)(x - 6) \\[1em] \Rightarrow [x + (-3)][x + (-6)] \\[1em] \Rightarrow x^2 + [(-3) + (-6)]x + (-3) \times (-6) \\[1em] \Rightarrow x^2 - 9x + 18\end{aligned}$$

Hence, $\bm{(x - 3)(x - 6) = x^2 - 9x + 18}$.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (z - 11)(z - 4) \\[1em] \Rightarrow [z + (-11)][z + (-4)] \\[1em] \Rightarrow z^2 + [(-11) + (-4)]z + (-11) \times (-4) \\[1em] \Rightarrow z^2 - 15z + 44\end{aligned}$$

Hence, $\bm{(z - 11)(z - 4) = z^2 - 15z + 44}$.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (b - 6)(b - 8) \\[1em] \Rightarrow [b + (-6)][b + (-8)] \\[1em] \Rightarrow b^2 + [(-6) + (-8)]b + (-6) \times (-8) \\[1em] \Rightarrow b^2 - 14b + 48\end{aligned}$$

Hence, $\bm{(b - 6)(b - 8) = b^2 - 14b + 48}$.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow \Big(a - \frac{3}{5}\Big)\Big(a - \frac{1}{3}\Big) \\[1em] \Rightarrow \Big[a + \Big(-\frac{3}{5}\Big)\Big]\Big[a + \Big(-\frac{1}{3}\Big)\Big] \\[1em] \Rightarrow a^2 + \Big[\Big(-\frac{3}{5}\Big) + \Big(-\frac{1}{3}\Big)\Big]a + \Big(-\frac{3}{5}\Big) \times \Big(-\frac{1}{3}\Big) \\[1em] \Rightarrow a^2 - \frac{9 + 5}{15}a + \frac{3}{15} \\[1em] \Rightarrow a^2 - \frac{14}{15}a + \frac{1}{5}\end{aligned}$$

Hence, $\bm{\Big(a - \frac{3}{5}\Big)\Big(a - \frac{1}{3}\Big) = a^2 - \frac{14}{15}a + \frac{1}{5}}$.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

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

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

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

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

Hence, $\bm{(p - 3)\Big(p + \frac{1}{2}\Big) = p^2 - \frac{5}{2}p - \frac{3}{2}}$.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (4p + 3)(4p + 7) \\[1em] \Rightarrow (4p)^2 + (3 + 7)(4p) + 3 \times 7 \\[1em] \Rightarrow 16p^2 + 10 \times 4p + 21 \\[1em] \Rightarrow 16p^2 + 40p + 21\end{aligned}$$

Hence, $\bm{(4p + 3)(4p + 7) = 16p^2 + 40p + 21}$.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (9c + 4)(9c - 2) \\[1em] \Rightarrow (9c + 4)[9c + (-2)] \\[1em] \Rightarrow (9c)^2 + \lbrack4 + (-2)\rbrack\lparen9c\rparen + 4 \times (-2) \\[1em] \Rightarrow 81c^2 + 2 \times 9c - 8 \\[1em] \Rightarrow 81c^2 + 18c - 8\end{aligned}$$

Hence, $\bm{(9c + 4)(9c - 2) = 81c^2 + 18c - 8}$.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (3a - 8)(3a + 2) \\[1em] \Rightarrow [3a + (-8)](3a + 2) \\[1em] \Rightarrow (3a)^2 +\lbrack-8 + 2\rbrack\lparen3a\rparen + (-8) \times 2 \\[1em] \Rightarrow 9a^2 - 6 \times 3a - 16 \\[1em] \Rightarrow 9a^2 - 18a - 16\end{aligned}$$

Hence, $\bm{(3a - 8)(3a + 2) = 9a^2 - 18a - 16}$.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (5x - 2)(5x - 7) \\[1em] \Rightarrow [5x + (-2)][5x + (-7)] \\[1em] \Rightarrow (5x)^2 + \lbrack(-2) + (-7)\rbrack\lparen5x\rparen + (-2) \times (-7) \\[1em] \Rightarrow 25x^2 - 9 \times 5x + 14 \\[1em] \Rightarrow 25x^2 - 45x + 14\end{aligned}$$

Hence, $\bm{(5x - 2)(5x - 7) = 25x^2 - 45x + 14}$.

Let $x^2 = y$, then the given product becomes $(y + 3)(y + 6)$

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (y + 3)(y + 6) \\[1em] \Rightarrow y^2 + (3 + 6)y + 3 \times 6 \\[1em] \Rightarrow y^2 + 9y + 18\end{aligned}$$

Substituting $y = x^2$, we get

$$\begin{aligned}\Rightarrow (x^2)^2 + 9x^2 + 18 \\[1em] \Rightarrow x^4 + 9x^2 + 18\end{aligned}$$

Hence, $\bm{(x^2 + 3)(x^2 + 6) = x^4 + 9x^2 + 18}$.

Let $y^2 = t$, then the given product becomes $(t - 1)(t + 4)$

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (t - 1)(t + 4) \\[1em] \Rightarrow [t + (-1)](t + 4) \\[1em] \Rightarrow t^2 + [(-1) + 4]t + (-1) \times 4 \\[1em] \Rightarrow t^2 + 3t - 4\end{aligned}$$

Substituting $t = y^2$, we get

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

Hence, $\bm{(y^2 - 1)(y^2 + 4) = y^4 + 3y^2 - 4}$.

Let $z^2 = y$, then the given product becomes $(y + 3)(y - 7)$

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

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

Substituting $y = z^2$, we get

$$\begin{aligned}\Rightarrow (z^2)^2 - 4z^2 - 21 \\[1em] \Rightarrow z^4 - 4z^2 - 21\end{aligned}$$

Hence, $\bm{(z^2 + 3)(z^2 - 7) = z^4 - 4z^2 - 21}$.

Let $t^2 = y$, then the given product becomes $(y - 2)(y - 5)$

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (y - 2)(y - 5) \\[1em] \Rightarrow [y + (-2)][y + (-5)] \\[1em] \Rightarrow y^2 + [(-2) + (-5)]y + (-2) \times (-5) \\[1em] \Rightarrow y^2 - 7y + 10\end{aligned}$$

Substituting $y = t^2$, we get

$$\begin{aligned}\Rightarrow (t^2)^2 - 7t^2 + 10 \\[1em] \Rightarrow t^4 - 7t^2 + 10\end{aligned}$$

Hence, $\bm{(t^2 - 2)(t^2 - 5) = t^4 - 7t^2 + 10}$.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (ab - 2)(ab + 4) \\[1em] \Rightarrow [ab + (-2)](ab + 4) \\[1em] \Rightarrow (ab)^2 + \lbrack(-2) + 4\rbrack \lparen ab \rparen + (-2) \times 4 \\[1em] \Rightarrow (ab)^2 + 2(ab) - 8 \\[1em] \Rightarrow a^2b^2 + 2ab - 8\end{aligned}$$

Hence, $\bm{(ab - 2)(ab + 4) = a^2b^2 + 2ab - 8}$.

Using distributive laws, we multiply each term of the first polynomial by each term of the second polynomial and then combine the like terms.

$$\begin{aligned}\Rightarrow (2 + xy)(3 - xy) \\[1em] \Rightarrow 2(3 - xy) + xy(3 - xy) \\[1em] \Rightarrow 6 - 2xy + 3xy - x^2y^2 \\[1em] \Rightarrow 6 + xy - x^2y^2\end{aligned}$$

Hence, $\bm{(2 + xy)(3 - xy) = 6 + xy - x^2y^2}$.

Using distributive laws, we multiply each term of the first polynomial by each term of the second polynomial and then combine the like terms.

$$\begin{aligned}\Rightarrow (3x + 4y)(4x + 3y) \\[1em] \Rightarrow 3x(4x + 3y) + 4y(4x + 3y) \\[1em] \Rightarrow 12x^2 + 9xy + 16xy + 12y^2 \\[1em] \Rightarrow 12x^2 + 25xy + 12y^2\end{aligned}$$

Hence, $\bm{(3x + 4y)(4x + 3y) = 12x^2 + 25xy + 12y^2}$.

Using distributive laws, we multiply each term of the first polynomial by each term of the second polynomial and then combine the like terms.

$$\begin{aligned}\Rightarrow (4a - 5b)(3a + 2b) \\[1em] \Rightarrow 4a(3a + 2b) - 5b(3a + 2b) \\[1em] \Rightarrow 12a^2 + 8ab - 15ab - 10b^2 \\[1em] \Rightarrow 12a^2 - 7ab - 10b^2\end{aligned}$$

Hence, $\bm{(4a - 5b)(3a + 2b) = 12a^2 - 7ab - 10b^2}$.

Using distributive laws, we multiply each term of the first polynomial by each term of the second polynomial and then combine the like terms.

$$\begin{aligned}\Rightarrow (2y + z)(7z - 3y) \\[1em] \Rightarrow 2y(7z - 3y) + z(7z - 3y) \\[1em] \Rightarrow 14yz - 6y^2 + 7z^2 - 3yz \\[1em] \Rightarrow 7z^2 - 6y^2 + 11yz\end{aligned}$$

Hence, $\bm{(2y + z)(7z - 3y) = 7z^2 - 6y^2 + 11yz}$.

Using distributive laws, we multiply each term of the first polynomial by each term of the second polynomial and then combine the like terms.

$$\begin{aligned}\Rightarrow (2m - 4n)(4m - 3n) \\[1em] \Rightarrow 2m(4m - 3n) - 4n(4m - 3n) \\[1em] \Rightarrow 8m^2 - 6mn - 16mn + 12n^2 \\[1em] \Rightarrow 8m^2 - 22mn + 12n^2\end{aligned}$$

Hence, $\bm{(2m - 4n)(4m - 3n) = 8m^2 - 22mn + 12n^2}$.

Using distributive laws, we multiply each term of the first polynomial by each term of the second polynomial and then combine the like terms.

$$\begin{aligned}\Rightarrow (2pq + 0.1mn)(0.2pq + 3mn) \\[1em] \Rightarrow 2pq(0.2pq + 3mn) + 0.1mn(0.2pq + 3mn) \\[1em] \Rightarrow 0.4p^2q^2 + 6pqmn + 0.02pqmn + 0.3m^2n^2 \\[1em] \Rightarrow 0.4p^2q^2 + (6 + 0.02)pqmn + 0.3m^2n^2 \\[1em] \Rightarrow 0.4p^2q^2 + 6.02pqmn + 0.3m^2n^2\end{aligned}$$

Hence, $\bm{(2pq + 0.1mn)(0.2pq + 3mn) = 0.4p^2q^2 + 6.02pqmn + 0.3m^2n^2}$.

Using distributive laws, we multiply each term of the first polynomial by each term of the second polynomial and then combine the like terms.

$$\begin{aligned}\Rightarrow \Big(\frac{4m}{p} - \frac{0.2n}{q}\Big)\Big(\frac{3m}{p} + \frac{0.5n}{q}\Big) \\[1em] \Rightarrow \frac{4m}{p}\Big(\frac{3m}{p} + \frac{0.5n}{q}\Big) - \frac{0.2n}{q}\Big(\frac{3m}{p} + \frac{0.5n}{q}\Big) \\[1em] \Rightarrow \frac{12m^2}{p^2} + \frac{2mn}{pq} - \frac{0.6mn}{pq} - \frac{0.1n^2}{q^2} \\[1em] \Rightarrow \frac{12m^2}{p^2} + (2 - 0.6)\frac{mn}{pq} - \frac{0.1n^2}{q^2} \\[1em] \Rightarrow \frac{12m^2}{p^2} + \frac{1.4mn}{pq} - \frac{0.1n^2}{q^2}\end{aligned}$$

Hence, $\bm{\Big(\frac{4m}{p} - \frac{0.2n}{q}\Big)\Big(\frac{3m}{p} + \frac{0.5n}{q}\Big) = \frac{12m^2}{p^2} + \frac{1.4mn}{pq} - \frac{0.1n^2}{q^2}}$.

Using distributive laws, we multiply each term of the first polynomial by each term of the second polynomial and then combine the like terms.

$$\begin{aligned}\Rightarrow \Big(\frac{3}{4}x - 2pq\Big)\Big(3x - \frac{4}{5}pq\Big) \\[1em] \Rightarrow \frac{3}{4}x\Big(3x - \frac{4}{5}pq\Big) - 2pq\Big(3x - \frac{4}{5}pq\Big) \\[1em] \Rightarrow \frac{9}{4}x^2 - \frac{3}{5}pqx - 6pqx + \frac{8}{5}p^2q^2 \\[1em] \Rightarrow \frac{9}{4}x^2 - \Big(\frac{3}{5} + 6\Big)pqx + \frac{8}{5}p^2q^2 \\[1em] \Rightarrow \frac{9}{4}x^2 - \frac{3 + 30}{5}pqx + \frac{8}{5}p^2q^2 \\[1em] \Rightarrow \frac{9}{4}x^2 - \frac{33}{5}pqx + \frac{8}{5}p^2q^2\end{aligned}$$

Hence, $\bm{\Big(\frac{3}{4}x - 2pq\Big)\Big(3x - \frac{4}{5}pq\Big) = \frac{9}{4}x^2 - \frac{33}{5}pqx + \frac{8}{5}p^2q^2}$.

Using distributive laws, we multiply each term of the first polynomial by each term of the second polynomial and then combine the like terms.

$$\begin{aligned}\Rightarrow (2a^2 + 3b^2)(3a^2 + 2b^2) \\[1em] \Rightarrow 2a^2(3a^2 + 2b^2) + 3b^2(3a^2 + 2b^2) \\[1em] \Rightarrow 6a^4 + 4a^2b^2 + 9a^2b^2 + 6b^4 \\[1em] \Rightarrow 6a^4 + 13a^2b^2 + 6b^4\end{aligned}$$

Hence, $\bm{(2a^2 + 3b^2)(3a^2 + 2b^2) = 6a^4 + 13a^2b^2 + 6b^4}$.

Using distributive laws, we multiply each term of the first polynomial by each term of the second polynomial and then combine the like terms.

$$\begin{aligned}\Rightarrow (x^2 + 5y^2)(y^2 - 2x^2) \\[1em] \Rightarrow x^2(y^2 - 2x^2) + 5y^2(y^2 - 2x^2) \\[1em] \Rightarrow x^2y^2 - 2x^4 + 5y^4 - 10x^2y^2 \\[1em] \Rightarrow -2x^4 + 5y^4 - 9x^2y^2\end{aligned}$$

Hence, $\bm{(x^2 + 5y^2)(y^2 - 2x^2) = -2x^4 + 5y^4 - 9x^2y^2}$.

Using distributive laws, we multiply each term of the first polynomial by each term of the second polynomial and then combine the like terms.

$$\begin{aligned}\Rightarrow (3c^2 - 4d^2)(4c^2 - 3d^2) \\[1em] \Rightarrow 3c^2(4c^2 - 3d^2) - 4d^2(4c^2 - 3d^2) \\[1em] \Rightarrow 12c^4 - 9c^2d^2 - 16c^2d^2 + 12d^4 \\[1em] \Rightarrow 12c^4 - 25c^2d^2 + 12d^4\end{aligned}$$

Hence, $\bm{(3c^2 - 4d^2)(4c^2 - 3d^2) = 12c^4 - 25c^2d^2 + 12d^4}$.

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

$$\begin{aligned}\Rightarrow (y + 9)(y - 9) \\[1em] \Rightarrow (y)^2 - (9)^2 \\[1em] \Rightarrow y^2 - 81\end{aligned}$$

Hence, $\bm{(y + 9)(y - 9) = y^2 - 81}$.

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

$$\begin{aligned}\Rightarrow (4 + b)(4 - b) \\[1em] \Rightarrow (4)^2 - (b)^2 \\[1em] \Rightarrow 16 - b^2\end{aligned}$$

Hence, $\bm{(4 + b)(4 - b) = 16 - b^2}$.

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

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

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

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

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

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

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

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

Hence, $\bm{(3x - 5)(3x + 5) = 9x^2 - 25}$.

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

$$\begin{aligned}\Rightarrow (2 + 7x)(2 - 7x) \\[1em] \Rightarrow (2)^2 - (7x)^2 \\[1em] \Rightarrow 4 - 49x^2\end{aligned}$$

Hence, $\bm{(2 + 7x)(2 - 7x) = 4 - 49x^2}$.

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

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

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

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

$$\begin{aligned}\Rightarrow (4x + 3y)(4x - 3y) \\[1em] \Rightarrow (4x)^2 - (3y)^2 \\[1em] \Rightarrow 16x^2 - 9y^2\end{aligned}$$

Hence, $\bm{(4x + 3y)(4x - 3y) = 16x^2 - 9y^2}$.

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

$$\begin{aligned}\Rightarrow \Big(\frac{a}{3} - \frac{b}{4}\Big)\Big(\frac{a}{3} + \frac{b}{4}\Big) \\[1em] \Rightarrow \Big(\frac{a}{3}\Big)^2 - \Big(\frac{b}{4}\Big)^2 \\[1em] \Rightarrow \frac{a^2}{9} - \frac{b^2}{16}\end{aligned}$$

Hence, $\bm{\Big(\frac{a}{3} - \frac{b}{4}\Big)\Big(\frac{a}{3} + \frac{b}{4}\Big) = \frac{a^2}{9} - \frac{b^2}{16}}$.

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

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

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

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

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

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

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

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

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

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

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

Hence, $\bm{\Big(\frac{1}{3x} + \frac{2}{5y}\Big)\Big(\frac{1}{3x} - \frac{2}{5y}\Big) = \frac{1}{9x^2} - \frac{4}{25y^2}}$.

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

$$\begin{aligned}\Rightarrow (1.1x - 0.3y)(1.1x + 0.3y) \\[1em] \Rightarrow (1.1x)^2 - (0.3y)^2 \\[1em] \Rightarrow 1.21x^2 - 0.09y^2\end{aligned}$$

Hence, $\bm{(1.1x - 0.3y)(1.1x + 0.3y) = 1.21x^2 - 0.09y^2}$.

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

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

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

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

$$\begin{aligned}\Rightarrow (6x^2 - 7y^2)(6x^2 + 7y^2) \\[1em] \Rightarrow (6x^2)^2 - (7y^2)^2 \\[1em] \Rightarrow 36x^4 - 49y^4\end{aligned}$$

Hence, $\bm{(6x^2 - 7y^2)(6x^2 + 7y^2) = 36x^4 - 49y^4}$.

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

$$\begin{aligned}\Rightarrow (4x^2 + 2yz)(2x^2 - yz) \\[1em] \Rightarrow 2(2x^2 + yz)(2x^2 - yz) \\[1em] \Rightarrow 2[(2x^2)^2 - (yz)^2] \\[1em] \Rightarrow 2(4x^4 - y^2z^2) \\[1em] \Rightarrow 8x^4 - 2y^2z^2\end{aligned}$$

Hence, $\bm{(4x^2 + 2yz)(2x^2 - yz) = 8x^4 - 2y^2z^2}$.

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

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

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

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

$$\begin{aligned}\Rightarrow (2x + 3)(2x - 3)(4x^2 + 9) \\[1em] \Rightarrow [(2x)^2 - (3)^2](4x^2 + 9) \\[1em] \Rightarrow (4x^2 - 9)(4x^2 + 9) \\[1em] \Rightarrow (4x^2)^2 - (9)^2 \\[1em] \Rightarrow 16x^4 - 81\end{aligned}$$

Hence, $\bm{(2x + 3)(2x - 3)(4x^2 + 9) = 16x^4 - 81}$.

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

$$\begin{aligned}\Rightarrow (x + 2y)(x - 2y)(x^2 + 4y^2) \\[1em] \Rightarrow [(x)^2 - (2y)^2](x^2 + 4y^2) \\[1em] \Rightarrow (x^2 - 4y^2)(x^2 + 4y^2) \\[1em] \Rightarrow (x^2)^2 - (4y^2)^2 \\[1em] \Rightarrow x^4 - 16y^4\end{aligned}$$

Hence, $\bm{(x + 2y)(x - 2y)(x^2 + 4y^2) = x^4 - 16y^4}$.

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

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

Hence, $\bm{(a + bc)(a - bc)(a^2 + b^2c^2) = a^4 - b^4c^4}$.

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

$$\begin{aligned}\Rightarrow \Big(\frac{2}{5} + x\Big)\Big(\frac{2}{5} - x\Big)\Big(\frac{4}{25} + x^2\Big) \\[1em] \Rightarrow \Big[\Big(\frac{2}{5}\Big)^2 - (x)^2\Big]\Big(\frac{4}{25} + x^2\Big) \\[1em] \Rightarrow \Big(\frac{4}{25} - x^2\Big)\Big(\frac{4}{25} + x^2\Big) \\[1em] \Rightarrow \Big(\frac{4}{25}\Big)^2 - (x^2)^2 \\[1em] \Rightarrow \frac{16}{625} - x^4\end{aligned}$$

Hence, $\bm{\Big(\frac{2}{5} + x\Big)\Big(\frac{2}{5} - x\Big)\Big(\frac{4}{25} + x^2\Big) = \frac{16}{625} - x^4}$.

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

$$\begin{aligned}\Rightarrow 88 \times 112 \\[1em] \Rightarrow (100 - 12)(100 + 12) \\[1em] \Rightarrow (100)^2 - (12)^2 \\[1em] \Rightarrow 10000 - 144 \\[1em] \Rightarrow 9856\end{aligned}$$

Hence, 88 × 112 = 9856.

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

$$\begin{aligned}\Rightarrow 153 \times 167 \\[1em] \Rightarrow (160 - 7)(160 + 7) \\[1em] \Rightarrow (160)^2 - (7)^2 \\[1em] \Rightarrow 25600 - 49 \\[1em] \Rightarrow 25551\end{aligned}$$

Hence, 153 × 167 = 25551.

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

$$\begin{aligned}\Rightarrow 10.8 \times 9.2 \\[1em] \Rightarrow (10 + 0.8)(10 - 0.8) \\[1em] \Rightarrow (10)^2 - (0.8)^2 \\[1em] \Rightarrow 100 - 0.64 \\[1em] \Rightarrow 99.36\end{aligned}$$

Hence, 10.8 × 9.2 = 99.36.

Converting the mixed fractions into improper fractions and using the identity $(a + b)(a - b) = (a^2 - b^2)$,

$$\begin{aligned}\Rightarrow 3\frac{1}{3} \times 4\frac{2}{3} \\[1em] \Rightarrow \frac{10}{3} \times \frac{14}{3} \\[1em] \Rightarrow \frac{1}{9} \times (10 \times 14) \\[1em] \Rightarrow \frac{1}{9} \times (12 - 2)(12 + 2) \\[1em] \Rightarrow \frac{1}{9} \times [(12)^2 - (2)^2] \\[1em] \Rightarrow \frac{1}{9} \times (144 - 4) \\[1em] \Rightarrow \frac{140}{9} \\[1em] \Rightarrow 15\frac{5}{9}\end{aligned}$$

Hence, $\mathbf{3\frac{1}{3} \times 4\frac{2}{3} = 15\frac{5}{9}}$.

Converting the mixed fractions into improper fractions and using the identity $(a + b)(a - b) = (a^2 - b^2)$,

$$\begin{aligned}\Rightarrow 9\frac{1}{4} \times 15\frac{3}{4} \\[1em] \Rightarrow \frac{37}{4} \times \frac{63}{4} \\[1em] \Rightarrow \frac{1}{16} \times (37 \times 63) \\[1em] \Rightarrow \frac{1}{16} \times (50 - 13)(50 + 13) \\[1em] \Rightarrow \frac{1}{16} \times [(50)^2 - (13)^2] \\[1em] \Rightarrow \frac{1}{16} \times (2500 - 169) \\[1em] \Rightarrow \frac{2331}{16} \\[1em] \Rightarrow 145\frac{11}{16}\end{aligned}$$

Hence, $\mathbf{9\frac{1}{4} \times 15\frac{3}{4} = 145\frac{11}{16}}$.

Using the identity $(a + b)^2 = a^2 + b^2 + 2ab$,

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

Hence, $\bm{(x + 3)^2 = x^2 + 9 + 6x}$.

Using the identity $(a + b)^2 = a^2 + b^2 + 2ab$,

$$\begin{aligned}\Rightarrow (2a + 7)^2 \\[1em] \Rightarrow (2a)^2 + (7)^2 + 2 \times 2a \times 7 \\[1em] \Rightarrow 4a^2 + 49 + 28a\end{aligned}$$

Hence, $\bm{(2a + 7)^2 = 4a^2 + 49 + 28a}$.

Using the identity $(a + b)^2 = a^2 + b^2 + 2ab$,

$$\begin{aligned}\Rightarrow (8 + 3p)^2 \\[1em] \Rightarrow (8)^2 + (3p)^2 + 2 \times 8 \times 3p \\[1em] \Rightarrow 64 + 9p^2 + 48p\end{aligned}$$

Hence, $\bm{(8 + 3p)^2 = 64 + 9p^2 + 48p}$.

Using the identity $(a + b)^2 = a^2 + b^2 + 2ab$,

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

Hence, $\bm{(\sqrt{3}x + 2)^2 = 3x^2 + 4 + 4\sqrt{3}x}$.

Using the identity $(a + b)^2 = a^2 + b^2 + 2ab$,

$$\begin{aligned}\Rightarrow (4 + \sqrt{5}y)^2 \\[1em] \Rightarrow (4)^2 + (\sqrt{5}y)^2 + 2 \times 4 \times \sqrt{5}y \\[1em] \Rightarrow 16 + 5y^2 + 8\sqrt{5}y\end{aligned}$$

Hence, $\bm{(4 + \sqrt{5}y)^2 = 16 + 5y^2 + 8\sqrt{5}y}$.

Using the identity $(a + b)^2 = a^2 + b^2 + 2ab$,

$$\begin{aligned}\Rightarrow (6x + 11y)^2 \\[1em] \Rightarrow (6x)^2 + (11y)^2 + 2 \times 6x \times 11y \\[1em] \Rightarrow 36x^2 + 121y^2 + 132xy\end{aligned}$$

Hence, $\bm{(6x + 11y)^2 = 36x^2 + 121y^2 + 132xy}$.

Using the identity $(a + b)^2 = a^2 + b^2 + 2ab$,

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

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

Using the identity $(a + b)^2 = a^2 + b^2 + 2ab$,

$$\begin{aligned}\Rightarrow \Big(\frac{3a}{5} + \frac{5b}{3}\Big)^2 \\[1em] \Rightarrow \Big(\frac{3a}{5}\Big)^2 + \Big(\frac{5b}{3}\Big)^2 + 2 \times \frac{3a}{5} \times \frac{5b}{3} \\[1em] \Rightarrow \frac{9a^2}{25} + \frac{25b^2}{9} + 2ab\end{aligned}$$

Hence, $\bm{\Big(\frac{3a}{5} + \frac{5b}{3}\Big)^2 = \frac{9a^2}{25} + \frac{25b^2}{9} + 2ab}$.

Using the identity $(a - b)^2 = a^2 + b^2 - 2ab$,

$$\begin{aligned}\Rightarrow (x - 9)^2 \\[1em] \Rightarrow (x)^2 + (9)^2 - 2 \times x \times 9 \\[1em] \Rightarrow x^2 + 81 - 18x\end{aligned}$$

Hence, $\bm{(x - 9)^2 = x^2 + 81 - 18x}$.

Using the identity $(a - b)^2 = a^2 + b^2 - 2ab$,

$$\begin{aligned}\Rightarrow (6 - y)^2 \\[1em] \Rightarrow (6)^2 + (y)^2 - 2 \times 6 \times y \\[1em] \Rightarrow 36 + y^2 - 12y\end{aligned}$$

Hence, $\bm{(6 - y)^2 = 36 + y^2 - 12y}$.

Using the identity $(a - b)^2 = a^2 + b^2 - 2ab$,

$$\begin{aligned}\Rightarrow (3a - 2)^2 \\[1em] \Rightarrow (3a)^2 + (2)^2 - 2 \times 3a \times 2 \\[1em] \Rightarrow 9a^2 + 4 - 12a\end{aligned}$$

Hence, $\bm{(3a - 2)^2 = 9a^2 + 4 - 12a}$.

Using the identity $(a - b)^2 = a^2 + b^2 - 2ab$,

$$\begin{aligned}\Rightarrow (8y - 5z)^2 \\[1em] \Rightarrow (8y)^2 + (5z)^2 - 2 \times 8y \times 5z \\[1em] \Rightarrow 64y^2 + 25z^2 - 80yz\end{aligned}$$

Hence, $\bm{(8y - 5z)^2 = 64y^2 + 25z^2 - 80yz}$.

Using the identity $(a - b)^2 = a^2 + b^2 - 2ab$,

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

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

Using the identity $(a - b)^2 = a^2 + b^2 - 2ab$,

$$\begin{aligned}\Rightarrow \Big(2a - \frac{5}{2}\Big)^2 \\[1em] \Rightarrow (2a)^2 + \Big(\frac{5}{2}\Big)^2 - 2 \times 2a \times \frac{5}{2} \\[1em] \Rightarrow 4a^2 + \frac{25}{4} - 10a\end{aligned}$$

Hence, $\bm{\Big(2a - \frac{5}{2}\Big)^2 = 4a^2 + \frac{25}{4} - 10a}$.

Using the identity $(a - b)^2 = a^2 + b^2 - 2ab$,

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

Hence, $\bm{\Big(\frac{2}{a} - \frac{3}{b}\Big)^2 = \frac{4}{a^2} + \frac{9}{b^2} - \frac{12}{ab}}$.

Using the identity $(a - b)^2 = a^2 + b^2 - 2ab$,

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

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

Using the identity $(a + b)^2 = a^2 + b^2 + 2ab$,

⇒ (53)²

⇒ (50 + 3)²

⇒ (50)² + (3)² + 2 × 50 × 3

⇒ 2500 + 9 + 300

⇒ 2809

Hence, (53)² = 2809.

Using the identity $(a + b)^2 = a^2 + b^2 + 2ab$,

⇒ (84)²

⇒ (80 + 4)²

⇒ (80)² + (4)² + 2 × 80 × 4

⇒ 6400 + 16 + 640

⇒ 7056

Hence, (84)² = 7056.

Using the identity $(a + b)^2 = a^2 + b^2 + 2ab$,

⇒ (1011)²

⇒ (1000 + 11)²

⇒ (1000)² + (11)² + 2 × 1000 × 11

⇒ 1000000 + 121 + 22000

⇒ 1022121

Hence, (1011)² = 1022121.

Using the identity $(a + b)^2 = a^2 + b^2 + 2ab$,

⇒ (10.9)²

⇒ (10 + 0.9)²

⇒ (10)² + (0.9)² + 2 × 10 × 0.9

⇒ 100 + 0.81 + 18

⇒ 118.81

Hence, (10.9)² = 118.81.

Using the identity $(a - b)^2 = a^2 + b^2 - 2ab$,

⇒ (67)²

⇒ (70 - 3)²

⇒ (70)² + (3)² - 2 × 70 × 3

⇒ 4900 + 9 - 420

⇒ 4489

Hence, (67)² = 4489.

Using the identity $(a - b)^2 = a^2 + b^2 - 2ab$,

⇒ (795)²

⇒ (800 - 5)²

⇒ (800)² + (5)² - 2 × 800 × 5

⇒ 640000 + 25 - 8000

⇒ 632025

Hence, (795)² = 632025.

Using the identity $(a - b)^2 = a^2 + b^2 - 2ab$,

⇒ (988)²

⇒ (1000 - 12)²

⇒ (1000)² + (12)² - 2 × 1000 × 12

⇒ 1000000 + 144 - 24000

⇒ 976144

Hence, (988)² = 976144.

Using the identity $(a - b)^2 = a^2 + b^2 - 2ab$,

⇒ (9.2)²

⇒ (10 - 0.8)²

⇒ (10)² + (0.8)² - 2 × 10 × 0.8

⇒ 100 + 0.64 - 16

⇒ 84.64

Hence, (9.2)² = 84.64.

Given,

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

(i) Squaring both sides,

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

Subtracting 4 from both sides,

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

Using the identity $\Big(a - \frac{1}{a}\Big)^2 = a^2 + \frac{1}{a^2} - 2$,

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

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

(ii) From part (i),

$x^2 + \frac{1}{x^2} + 2 = 16$

Subtracting 2 from both sides,

$$\begin{aligned}\Rightarrow x^2 + \frac{1}{x^2} = 16 - 2 \\[1em] \Rightarrow x^2 + \frac{1}{x^2} = 14\end{aligned}$$

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

(iii) From part (ii),

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

Squaring both sides,

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

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

Given,

$z - \frac{1}{z} = 6$

(i) Squaring both sides and using the identity $\Big(a - \frac{1}{a}\Big)^2 = a^2 + \frac{1}{a^2} - 2$,

$$\begin{aligned}\Rightarrow \Big(z - \frac{1}{z}\Big)^2 = (6)^2 \\[1em] \Rightarrow z^2 + \frac{1}{z^2} - 2 = 36\end{aligned}$$

Adding 4 on both sides,

$$\begin{aligned}\Rightarrow \Big(z^2 + \frac{1}{z^2} - 2\Big) + 4 = 36 + 4 \\[1em] \Rightarrow z^2 + \frac{1}{z^2} + 2 = 40\end{aligned}$$

Using the identity $\Big(a + \frac{1}{a}\Big)^2 = a^2 + \frac{1}{a^2} + 2$,

$$\begin{aligned}\Rightarrow \Big(z + \frac{1}{z}\Big)^2 = 40 \\[1em] \Rightarrow z + \frac{1}{z} = \pm\sqrt{40} \\[1em] \Rightarrow z + \frac{1}{z} = \pm 2\sqrt{10}\end{aligned}$$

Hence, $\bm{z + \frac{1}{z} = \pm 2\sqrt{10}}$.

(ii) From part (i),

$z^2 + \frac{1}{z^2} - 2 = 36$

Adding 2 on both sides,

$$\begin{aligned}\Rightarrow z^2 + \frac{1}{z^2} = 36 + 2 \\[1em] \Rightarrow z^2 + \frac{1}{z^2} = 38\end{aligned}$$

Hence, $\bm{z^2 + \frac{1}{z^2} = 38}$.

(iii) From part (ii),

$z^2 + \frac{1}{z^2} = 38$

Squaring both sides,

$$\begin{aligned}\Rightarrow \Big(z^2 + \frac{1}{z^2}\Big)^2 = (38)^2 \\[1em] \Rightarrow z^4 + \frac{1}{z^4} + 2 = 1444 \\[1em] \Rightarrow z^4 + \frac{1}{z^4} = 1444 - 2 \\[1em] \Rightarrow z^4 + \frac{1}{z^4} = 1442\end{aligned}$$

Hence, $\bm{z^4 + \frac{1}{z^4} = 1442}$.

Given,

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

Adding 2 on both sides,

$$\begin{aligned}\Rightarrow a^2 + \frac{1}{a^2} + 2 = 23 + 2 \\[1em] \Rightarrow a^2 + \frac{1}{a^2} + 2 = 25\end{aligned}$$

Using the identity $\Big(a + \frac{1}{a}\Big)^2 = a^2 + \frac{1}{a^2} + 2$,

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

Hence, $\bm{a + \frac{1}{a} = \pm 5}$.

Given,

$x^2 + \frac{1}{x^2} = 102$

Subtracting 2 from both sides,

$$\begin{aligned}\Rightarrow x^2 + \frac{1}{x^2} - 2 = 102 - 2 \\[1em] \Rightarrow x^2 + \frac{1}{x^2} - 2 = 100\end{aligned}$$

Using the identity $\Big(a - \frac{1}{a}\Big)^2 = a^2 + \frac{1}{a^2} - 2$,

$$\begin{aligned}\Rightarrow \Big(x - \frac{1}{x}\Big)^2 = 100 \\[1em] \Rightarrow x - \frac{1}{x} = \pm\sqrt{100} \\[1em] \Rightarrow x - \frac{1}{x} = \pm 10\end{aligned}$$

Hence, $\bm{x - \frac{1}{x} = \pm 10}$.

Given,

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

Squaring both sides and using the identity $\Big(a + \frac{1}{a}\Big)^2 = a^2 + \frac{1}{a^2} + 2$,

$$\begin{aligned}\Rightarrow \Big(2p + \frac{1}{2p}\Big)^2 = (5)^2 \\[1em] \Rightarrow (2p)^2 + \Big(\frac{1}{2p}\Big)^2 + 2 = 25 \\[1em] \Rightarrow 4p^2 + \frac{1}{4p^2} + 2 = 25 \\[1em] \Rightarrow 4p^2 + \frac{1}{4p^2} = 25 - 2 \\[1em] \Rightarrow 4p^2 + \frac{1}{4p^2} = 23\end{aligned}$$

Hence, $\bm{4p^2 + \frac{1}{4p^2} = 23}$.

Given,

$3c - \frac{1}{3c} = 8$

Squaring both sides and using the identity $\Big(a - \frac{1}{a}\Big)^2 = a^2 + \frac{1}{a^2} - 2$,

$$\begin{aligned}\Rightarrow \Big(3c - \frac{1}{3c}\Big)^2 = (8)^2 \\[1em] \Rightarrow (3c)^2 + \Big(\frac{1}{3c}\Big)^2 - 2 = 64 \\[1em] \Rightarrow 9c^2 + \frac{1}{9c^2} - 2 = 64 \\[1em] \Rightarrow 9c^2 + \frac{1}{9c^2} = 64 + 2 \\[1em] \Rightarrow 9c^2 + \frac{1}{9c^2} = 66\end{aligned}$$

Hence, $\bm{9c^2 + \frac{1}{9c^2} = 66}$.

Given,

$a + b = 8$ and $ab = 15$

We know that,

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

Substituting the values,

$$\begin{aligned}\Rightarrow (8)^2 = a^2 + b^2 + 2 \times 15 \\[1em] \Rightarrow 64 = a^2 + b^2 + 30 \\[1em] \Rightarrow a^2 + b^2 = 64 - 30 \\[1em] \Rightarrow a^2 + b^2 = 34\end{aligned}$$

Hence, $\bm{a^2 + b^2 = 34}$.

Given,

$a + b = 11$ and $a^2 + b^2 = 61$

We know that,

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

Substituting the values,

$$\begin{aligned}\Rightarrow (11)^2 = 61 + 2ab \\[1em] \Rightarrow 121 = 61 + 2ab \\[1em] \Rightarrow 2ab = 121 - 61 \\[1em] \Rightarrow 2ab = 60 \\[1em] \Rightarrow ab = \frac{60}{2} \\[1em] \Rightarrow ab = 30\end{aligned}$$

Hence, $\bm{ab = 30}$.

Given,

$a^2 + b^2 = 13$ and $ab = 6$

We know that,

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

Substituting the values,

$$\begin{aligned}\Rightarrow (a + b)^2 = 13 + 2 \times 6 \\[1em] \Rightarrow (a + b)^2 = 13 + 12 \\[1em] \Rightarrow (a + b)^2 = 25 \\[1em] \Rightarrow a + b = \pm\sqrt{25} \\[1em] \Rightarrow a + b = \pm 5\end{aligned}$$

Hence, $\bm{a + b = \pm 5}$.

Given,

$a + b = 15$ and $ab = 56$

We know that,

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

Substituting the values,

$$\begin{aligned}\Rightarrow (15)^2 = a^2 + b^2 + 2 \times 56 \\[1em] \Rightarrow 225 = a^2 + b^2 + 112 \\[1em] \Rightarrow a^2 + b^2 = 225 - 112 \\[1em] \Rightarrow a^2 + b^2 = 113\end{aligned}$$

Hence, $\bm{a^2 + b^2 = 113}$.

Given,

$a - b = 1$ and $ab = 12$

We know that,

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

Substituting the values,

$$\begin{aligned}\Rightarrow (1)^2 = a^2 + b^2 - 2 \times 12 \\[1em] \Rightarrow 1 = a^2 + b^2 - 24 \\[1em] \Rightarrow a^2 + b^2 = 1 + 24 \\[1em] \Rightarrow a^2 + b^2 = 25\end{aligned}$$

Hence, $\bm{a^2 + b^2 = 25}$.

Given,

$a - b = 5$ and $a^2 + b^2 = 53$

We know that,

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

Substituting the values,

$$\begin{aligned}\Rightarrow (5)^2 = 53 - 2ab \\[1em] \Rightarrow 25 = 53 - 2ab \\[1em] \Rightarrow 2ab = 53 - 25 \\[1em] \Rightarrow 2ab = 28 \\[1em] \Rightarrow ab = \frac{28}{2} \\[1em] \Rightarrow ab = 14\end{aligned}$$

Hence, $\bm{ab = 14}$.

Given,

$a^2 + b^2 = 52$ and $ab = 24$

We know that,

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

Substituting the values,

$$\begin{aligned}\Rightarrow (a - b)^2 = 52 - 2 \times 24 \\[1em] \Rightarrow (a - b)^2 = 52 - 48 \\[1em] \Rightarrow (a - b)^2 = 4 \\[1em] \Rightarrow a - b = \pm\sqrt{4} \\[1em] \Rightarrow a - b = \pm 2\end{aligned}$$

Hence, $\bm{a - b = \pm 2}$.

The given expression is of the form $a^2 + b^2 + 2ab$, which is the expansion of $(a + b)^2$.

$$\begin{aligned}\Rightarrow 36x^2 + 49y^2 + 84xy \\[1em] \Rightarrow (6x)^2 + (7y)^2 + 2 \times 6x \times 7y \\[1em] \Rightarrow (6x + 7y)^2\end{aligned}$$

Substituting $x = 3$ and $y = 6$,

$$\begin{aligned}\Rightarrow (6 \times 3 + 7 \times 6)^2 \\[1em] \Rightarrow (18 + 42)^2 \\[1em] \Rightarrow (60)^2 \\[1em] \Rightarrow 3600\end{aligned}$$

Hence, the value of the expression = 3600.

The given expression is of the form $a^2 + b^2 - 2ab$, which is the expansion of $(a - b)^2$.

$$\begin{aligned}\Rightarrow 25x^2 + 16y^2 - 40xy \\[1em] \Rightarrow (5x)^2 + (4y)^2 - 2 \times 5x \times 4y \\[1em] \Rightarrow (5x - 4y)^2\end{aligned}$$

Substituting $x = 6$ and $y = 7$,

$$\begin{aligned}\Rightarrow (5 \times 6 - 4 \times 7)^2 \\[1em] \Rightarrow (30 - 28)^2 \\[1em] \Rightarrow (2)^2 \\[1em] \Rightarrow 4\end{aligned}$$

Hence, the value of the expression = 4.

Given,

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

Squaring both sides and using the identity $\Big(a + \frac{1}{a}\Big)^2 = a^2 + \frac{1}{a^2} + 2$,

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

Hence, option 3 is the correct option.

Given,

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

Squaring both sides and using the identity $\Big(a + \frac{1}{a}\Big)^2 = a^2 + \frac{1}{a^2} + 2$,

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{x}\Big)^2 = (4)^2 \\[1em] \Rightarrow x^2 + \frac{1}{x^2} + 2 = 16 \\[1em] \Rightarrow x^2 + \frac{1}{x^2} = 14\end{aligned}$$

Squaring both sides again,

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

Hence, option 3 is the correct option.

Given,

$x^2 + \frac{1}{x^2} = 102$

Subtracting 2 from both sides,

$$\begin{aligned}\Rightarrow x^2 + \frac{1}{x^2} - 2 = 102 - 2 \\[1em] \Rightarrow x^2 + \frac{1}{x^2} - 2 = 100\end{aligned}$$

Using the identity $\Big(a - \frac{1}{a}\Big)^2 = a^2 + \frac{1}{a^2} - 2$,

$$\begin{aligned}\Rightarrow \Big(x - \frac{1}{x}\Big)^2 = 100 \\[1em] \Rightarrow x - \frac{1}{x} = \pm\sqrt{100} \\[1em] \Rightarrow x - \frac{1}{x} = \pm 10\end{aligned}$$

Since the options contain only the positive value, we take 10.

Hence, option 2 is the correct option.

Given,

$x + y = 7$ and $xy = 12$

We know that,

$(x + y)^2 = x^2 + y^2 + 2xy$

Substituting the values,

$$\begin{aligned}\Rightarrow (7)^2 = x^2 + y^2 + 2 \times 12 \\[1em] \Rightarrow 49 = x^2 + y^2 + 24 \\[1em] \Rightarrow x^2 + y^2 = 49 - 24 \\[1em] \Rightarrow x^2 + y^2 = 25\end{aligned}$$

Hence, option 1 is the correct option.

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

⇒ 107 × 93

⇒ (100 + 7)(100 - 7)

⇒ (100)² - (7)²

⇒ 10000 - 49

⇒ 9951

Hence, option 2 is the correct option.

Using the identity $a^2 - b^2 = (a + b)(a - b)$,

⇒ (370)² - (369)²

⇒ (370 + 369)(370 - 369)

⇒ 739 × 1

⇒ 739

Hence, option 3 is the correct option.

Given,

$a - b = 7$ and $ab = 9$

We know that,

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

Substituting the values,

$$\begin{aligned}\Rightarrow (7)^2 = a^2 + b^2 - 2 \times 9 \\[1em] \Rightarrow 49 = a^2 + b^2 - 18 \\[1em] \Rightarrow a^2 + b^2 = 49 + 18 \\[1em] \Rightarrow a^2 + b^2 = 67\end{aligned}$$

Hence, option 1 is the correct option.

A trinomial of the form $a^2 + b^2 + 2ab$ is a perfect square trinomial.

In the given expression, $4x^2 = (2x)^2$ and $20x = 2 \times 2x \times 5$, so here $a = 2x$ and $b = 5$.

So, the constant term of the perfect square trinomial must be $(5)^2 = 25$.

⇒ Required number = 25 - 16

⇒ 9

On adding 9, the expression becomes $4x^2 + 20x + 25 = (2x + 5)^2$, which is a perfect square.

Hence, option 1 is the correct option.

Given,

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

Squaring both sides and using the identity $\Big(a - \frac{1}{a}\Big)^2 = a^2 + \frac{1}{a^2} - 2$,

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

Hence, option 3 is the correct option.

Option 1:

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

So, it is equal to $a^2 + b^2$.

Option 2:

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

So, it is equal to $a^2 + b^2$.

Option 3:

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

So, it is not equal to $a^2 + b^2$.

Option 4:

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

So, it is equal to $a^2 + b^2$.

Hence, option 3 is the correct option.

Using the identities

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

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

Hence, option 2 is the correct option.

Using the identity $(a + b)^2 - (a - b)^2 = 4ab$,

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

Given,

$x + \frac{1}{x} = \frac{17}{4}$

Substituting values,

$$\begin{aligned}\Rightarrow \Big(x - \frac{1}{x}\Big)^2 = \Big(\frac{17}{4}\Big)^2 - 4 \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big)^2 = \frac{289}{16} - 4 \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big)^2 = \frac{289 - 64}{16} \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big)^2 = \frac{225}{16} \\[1em] \Rightarrow x - \frac{1}{x} = \pm\sqrt{\frac{225}{16}} \\[1em] \Rightarrow x - \frac{1}{x} = \pm \frac{15}{4}\end{aligned}$$

Since the options contain only the positive value, we take $\frac{15}{4}$.

Hence, option 4 is the correct option.

Expanding the square,

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

So, the Reason (R) is true.

Squaring $x + \frac{1}{x} = 4$ and using the Reason (R),

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{x}\Big)^2 = (4)^2 \\[1em] \Rightarrow x^2 + \frac{1}{x^2} + 2 = 16 \\[1em] \Rightarrow x^2 + \frac{1}{x^2} = 16 - 2 \\[1em] \Rightarrow x^2 + \frac{1}{x^2} = 14\end{aligned}$$

So, the Assertion (A) is true.

The Assertion (A) is obtained directly by using the identity stated in the Reason (R). So, the Reason (R) is the correct explanation of the Assertion (A).

Hence, option 1 is the correct option.

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$,

$$\begin{aligned}\Rightarrow (2x - 4)(2x + 1) \\[1em] \Rightarrow [2x + (-4)](2x + 1) \\[1em] \Rightarrow (2x)^2 + <a href="2x">(-4) + 1</a> + (-4) \times 1 \\[1em] \Rightarrow 4x^2 - 3(2x) - 4 \\[1em] \Rightarrow 4x^2 - 6x - 4\end{aligned}$$

The product is $4x^2 - 6x - 4$ and not $4x^2 - 3x - 4$. So, the Assertion (A) is false.

$(x - a)(x + a) = x^2 - a^2$ is the product of the sum and the difference of two terms, which is a correct identity. So, the Reason (R) is true.

Hence, option 4 is the correct option.

As 998 can be written as (1000 - 2), its square can be found by using the identity $(x - y)^2 = x^2 + y^2 - 2xy$. So, the Reason (R) is true.

Using this identity,

⇒ (998)²

⇒ (1000 - 2)²

⇒ (1000)² + (2)² - 2 × 1000 × 2

⇒ 1000000 + 4 - 4000

⇒ 996004

So, the Assertion (A) is true.

The Assertion (A) is obtained by using the identity stated in the Reason (R). So, the Reason (R) is the correct explanation of the Assertion (A).

Hence, option 1 is the correct option.

Using the identities $(a + b)^2 = a^2 + b^2 + 2ab$ and $(a - b)^2 = a^2 + b^2 - 2ab$,

$$\begin{aligned}\Rightarrow (3x + 2y)^2 - (3x - 2y)^2 \\[1em] \Rightarrow (9x^2 + 4y^2 + 12xy) - (9x^2 + 4y^2 - 12xy) \\[1em] \Rightarrow 9x^2 + 4y^2 + 12xy - 9x^2 - 4y^2 + 12xy \\[1em] \Rightarrow 24xy\end{aligned}$$

Hence, option 2 is the correct option.

Using the identity $a^2 - b^2 = (a + b)(a - b)$,

⇒ (9.7)² - (0.3)²

⇒ (9.7 + 0.3)(9.7 - 0.3)

⇒ 10 × 9.4

⇒ 94

Hence, option 4 is the correct option.

Using the identities $(a + b)^2 = a^2 + b^2 + 2ab$ and $(a - b)^2 = a^2 + b^2 - 2ab$,

$$\begin{aligned}\Rightarrow (3p + q)^2 - (3p - q)^2 \\[1em] \Rightarrow (9p^2 + q^2 + 6pq) - (9p^2 + q^2 - 6pq) \\[1em] \Rightarrow 9p^2 + q^2 + 6pq - 9p^2 - q^2 + 6pq \\[1em] \Rightarrow 12pq\end{aligned}$$

Comparing with the given expression,

$$\begin{aligned}\Rightarrow apq = 12pq \\[1em] \Rightarrow a = 12\end{aligned}$$

Hence, option 3 is the correct option.

Given,

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

Squaring both sides and using the identity $\Big(a - \frac{1}{a}\Big)^2 = a^2 + \frac{1}{a^2} - 2$,

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

Squaring both sides again,

$$\begin{aligned}\Rightarrow \Big(x^2 + \frac{1}{x^2}\Big)^2 = (11)^2 \\[1em] \Rightarrow x^4 + \frac{1}{x^4} + 2 = 121 \\[1em] \Rightarrow x^4 + \frac{1}{x^4} = 121 - 2 \\[1em] \Rightarrow x^4 + \frac{1}{x^4} = 119\end{aligned}$$

Hence, option 2 is the correct option.

Given,

$x^2 + \frac{1}{x^2} = 627$

Using the identity $\Big(a - \frac{1}{a}\Big)^2 = a^2 + \frac{1}{a^2} - 2$,

$$\begin{aligned}\Rightarrow \Big(x - \frac{1}{x}\Big)^2 = 627 - 2 \\[1em] \Rightarrow \Big(x - \frac{1}{x}\Big)^2 = 625 \\[1em] \Rightarrow x - \frac{1}{x} = \pm\sqrt{625} \\[1em] \Rightarrow x - \frac{1}{x} = \pm 25\end{aligned}$$

Since the options contain only the positive value, we take 25.

Hence, option 1 is the correct option.

Using the identity $a^2 - b^2 = (a + b)(a - b)$ to factorise the numerator,

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

Hence, option 4 is the correct option.

Using the identity $(a + b)(a - b) = (a^2 - b^2)$,

$$\begin{aligned}\Rightarrow 98 \times 102 \\[1em] \Rightarrow (100 - 2)(100 + 2) \\[1em] \Rightarrow (100)^2 - (2)^2\end{aligned}$$

Hence, option 2 is the correct option.