If $z=x-iy\text{ and }{z}^{1/3}=p+iq$, then $\left(\frac{x}{p}+\frac{y}{q}\right)/\left(p^2+q^2\right)$ is equal to

Given that,   $z^{1/3}=p+iq$

$\therefore z=(p+iq{)}^3=p^3-i{q}^3+3i{p}^{2}q-3p{q}^2$

and              $z=x-iy$                       (given)

$\begin{array}{lc}\therefore & x-iy=p^3-3p{q}^2+i\left(3p^{2}q-q^3\right)\\ \Rightarrow & x=p^3-3p{q}^2\\ \text{ and }& -y=3p^{2}q-q^3\\ \Rightarrow & \frac{x}{p}=p^2-3q^2\\ \text{ and }& \frac{y}{q}=-3p^2+q^2\\ \therefore & \frac{x}{p}+\frac{y}{q}=-2p^2-2q^2\\ \Rightarrow & \frac{1}{\left(p^2+q^2\right)}\left(\frac{x}{p}+\frac{y}{q}\right)=-2\end{array}$