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If $z=x$ - iy and $z^{1 / 3}=p+$ iq $(x, y, p, q \in \mathbb{R})$, then $\frac{\left(\frac{x}{p}+\frac{y}{q}\right)}{\left(p^2+q^2\right)}$ is equal to
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The correct answer is:
$-2$
$z=p^3-3 p q^2+i\left(3 p^2 q-q^3\right)=x-i y$
$\begin{aligned}
&x=p\left(p^2-3 q^2\right) \\
&y=-q\left(3 p^2-q^2\right) \\
&\therefore \frac{x}{p}+\frac{y}{q}=p^2-3 q^2-3 p^2+q^2=-2\left(p^2+q^2\right)
\end{aligned}$
$\begin{aligned}
&x=p\left(p^2-3 q^2\right) \\
&y=-q\left(3 p^2-q^2\right) \\
&\therefore \frac{x}{p}+\frac{y}{q}=p^2-3 q^2-3 p^2+q^2=-2\left(p^2+q^2\right)
\end{aligned}$
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