Prove that

Solution:

Given that, $(x+i y)^{1 / 3}=(a+i b)$

$\Rightarrow(x+i y)=(a+i b)^{3}$

$\Rightarrow(a+i b)^{3}=x+i y$

$\Rightarrow a^{3}+(i b)^{3}+3 a^{2} i b+3 a i^{2} b^{2}=x+i y$

$\Rightarrow a^{3}-i b^{3}+3 a^{2} i b-3 a b^{2}=x+i y$

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

On equating real and imaginary parts, we get

$x=a^{3}-3 a b^{2}$ and $y=3 a^{2} b-b^{3}$

Now, $\frac{x}{a}+\frac{y}{b}=\frac{\mathrm{a}^{3}-3 \mathrm{ab}^{2}}{a}+\frac{3 \mathrm{a}^{2} \mathrm{~b}-\mathrm{b}^{3}}{b}$

$=\frac{a\left(\mathrm{a}^{2}-3 \mathrm{~b}^{2}\right)}{a}+\frac{\mathrm{b}\left(3 \mathrm{a}^{2}-\mathrm{b}^{2}\right)}{b}$

$=a^{2}-3 b^{2}+3 a^{2}-b^{2}$

$=4 a^{2}-4 b^{2}$

$=4\left(a^{2}-b^{2}\right)$

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