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