Explanation

(b)
Given, $(x + iy)^{1/3} = a - ib$
And $\frac{x}{a} - \frac{y}{b} = k(a^{2} - b^{2})$
$\therefore\ \ \ x + iy = (a - ib)^{3} = \left( a^{3} - 3ab^{2} \right) + i(b^{3} - 3a^{2}b)$
$\therefore\ \ \ x = a^{3} - 3ab^{2},\ \ \ \ y = b^{3} - 3a^{2}b$
$\Rightarrow \ \frac{x}{a} = a^{2} - 3b^{2},\ \ \frac{y}{b} = b^{2} - 3a^{2}$
$\therefore\ \frac{x}{a} - \frac{y}{b} = a^{2} - 3b^{2} - b^{2} + 3a^{2} = 4(a^{2} - b^{2})$
But $\frac{x}{a} - \frac{y}{b} = k(a^{2} - b^{2})$ [given]
$\therefore\ \ \ \ \ k = 4$