Using standard formulae, expand each of the following:

(i) $\Big(a^2 - \dfrac{b}{2}\Big)^2$

(ii) $\Big(\dfrac{3a}{2b} - \dfrac{2b}{3a}\Big)^2$

(iii) $\Big(5x - \dfrac{2}{3x}\Big)^2$

We know that,

⇒ (a - b)² = a² + b² - 2ab.

(i) Given,

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

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

(ii) Given,

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

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

(iii) Given,

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

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