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If $(x+i y)^{3}=(u+i v)$ then prove that $\left(\frac{u}{x}+\frac{v}{y}\right)=4\left(x^{2}-y^{2}\right)$



Given that, $(x+i y)^{3}=(u+i v)$

$\Rightarrow x^{3}+(i y)^{3}+3 x^{2} y+3 x i^{2} y^{2}=u+i v$

$\Rightarrow x^{3}-i y^{3}+3 x^{2} y-3 x y^{2}=u+i v$

$\Rightarrow x^{3}-3 x y^{2}+i\left(3 x^{2} y-y^{3}\right)=u+i v$

On equating real and imaginary parts, we get

$U=x^{3}-3 x y^{2}$ and $v=3 x^{2} y-y^{3}$

Now, $\frac{u}{x}+\frac{v}{y}=\frac{\mathrm{x}^{3}-3 \mathrm{xy}^{2}}{x}+\frac{3 \mathrm{x}^{2} \mathrm{y}-\mathrm{y}^{3}}{y}$

$=\frac{x\left(\mathrm{x}^{2}-3 \mathrm{y}^{2}\right)}{x}+\frac{\mathrm{y}\left(3 \mathrm{x}^{2}-\mathrm{y}^{2}\right)}{y}$

$=x^{2}-3 y^{2}+3 x^{2}-y^{2}$

$=4 x^{2}-4 y^{2}$


Hence, $\frac{u}{x}+\frac{v}{y}=4\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right)$


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