If \(\mathrm{z}=x+\mathrm{i} y\) and \(\mathrm{z}^{1 / 3}=\mathrm{p}+\mathrm{iq}\), where \(x, y, \mathrm{p}\), \(\mathrm{q} \in \mathrm{R}\) and \(\mathrm{i}=\sqrt{-1}\), then value of \(\left(\frac{x}{\mathrm{p}}+\frac{y}{\mathrm{q}}\right)\) is

\(\mathrm{z}^{\frac{1}{3}}=\mathrm{p}+\mathrm{iq} \)
\( \Rightarrow \mathrm{z}=(\mathrm{p}+\mathrm{iq})^3 \)
\( \Rightarrow x+\mathrm{i} y=\mathrm{p}^3+3 \mathrm{p}^2 \mathrm{qi}+3 \mathrm{p}(\mathrm{iq})^2+(\mathrm{iq})^3 \)
\( \Rightarrow x+\mathrm{i} y=\left(\mathrm{p}^3-3 \mathrm{pq}^2\right)+\left(3 \mathrm{p}^2 \mathrm{q}-\mathrm{q}^3\right) \mathrm{i} \)
\( \Rightarrow x=\mathrm{p}^3-3 \mathrm{pq}^2 \text { and } y=3 \mathrm{p}^2 \mathrm{q}-\mathrm{q}^3 \)
\( \Rightarrow \frac{x}{\mathrm{p}}=\mathrm{p}^2-3 \mathrm{q}^2 \text { and } \frac{y}{\mathrm{q}}=3 \mathrm{p}^2-\mathrm{q}^2 \)
\( \therefore \left(\frac{x}{\mathrm{p}}+\frac{y}{\mathrm{q}}\right)=4 \mathrm{p}^2-4 \mathrm{q}^2=4\left(\mathrm{p}^2-\mathrm{q}^2\right)\)