Explanation

(d)
Given, $z^{1/3} = p + iq$
$\Rightarrow \ \ \ \ \ \ (x - iy) = (p + iq)^{3}$ [put $z = x - iy$]
$\Rightarrow \ \ \ \ \ \ (x - iy) = p^{3}—iq^{3} + 3p^{2}qi - 3pq^{2}$
$\Rightarrow \ \ \ (x - iy) = \left( p^{3} - 3pq^{2} \right) + i\left( 3p^{2}q - q^{3} \right)$
$\Rightarrow \ \ x = \left( p^{3} - 3pq^{2} \right)$ and $- y = 3p^{2}q - q^{3}$
$\Rightarrow \ \frac{x}{p} = (p^{2} - 3q^{2})$ and $\frac{y}{q} = (q^{2} - 3p^{2})$
$\therefore\ \ \frac{x}{p} + \frac{y}{q} = - 2p^{2} - 2q^{2}$
$\Rightarrow \frac{\frac{x}{p} + \frac{y}{q}}{(p^{2} + q^{2})} = - 2$