Using standard formulae, expand each of the following:

(i) $\Big(\dfrac{2}{5}x + \dfrac{5}{6}y\Big)^2$

(ii) $\Big(\dfrac{x}{3} + \dfrac{6}{x}\Big)^2$

(iii) $\Big(6 + \dfrac{5}{x}\Big)^2$

We know that,

⇒ (a + b)² = a² + b² + 2ab.

(i) Given,

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

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

(ii) Given,

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

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

(iii) Given,

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

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