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If $z=x+i y, z^{1 / 3}=a-i b$, then $\frac{x}{a}-\frac{y}{b}=k\left(a^{2}-\right.$ $\mathrm{b}^{2}$ ) where $\mathrm{k}$ is equal to
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4
$\mathrm{z}^{1 / 3}=\mathrm{a}-\mathrm{ib} \Rightarrow \mathrm{z}=(\mathrm{a}-\mathrm{ib})^{3}$ $\therefore x+i y=a^{3}+i b^{3}-3 i a^{2} b-3 a b^{2}$. Then
$$
\begin{array}{l}
x=a^{3}-3 a b^{2} \Rightarrow \frac{x}{a}=a^{2}-3 b^{2} \\
y=b^{3}-3 a^{2} b \Rightarrow \frac{y}{b}=b^{2}-3 a^{2}
\end{array}
$$
So, $\frac{x}{a}-\frac{y}{b}=4\left(a^{2}-b^{2}\right)$
$$
\begin{array}{l}
x=a^{3}-3 a b^{2} \Rightarrow \frac{x}{a}=a^{2}-3 b^{2} \\
y=b^{3}-3 a^{2} b \Rightarrow \frac{y}{b}=b^{2}-3 a^{2}
\end{array}
$$
So, $\frac{x}{a}-\frac{y}{b}=4\left(a^{2}-b^{2}\right)$
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