$\sqrt[3]{a+i b}=x+i y$ হলে,$-2\left(x^{2}+y^{2}\right)=?$

Solve: দেওয়া আছে , $\sqrt[3]{a+i b}=x+i y$

$\begin{aligned} \therefore \mathrm{a}+i \mathrm{~b} & =x^{3}+3 \cdot x^{2} \cdot \mathrm{i} y+3 \cdot x \cdot(i y)^{2}+(i y)^{3} \\ & =x^{3}+3 x^{2} y \mathrm{i}-3 x y^{2}-i y^{3} \end{aligned}$

উভয় পক্ষ হতে বাষ্তব ও কাল্পনিক অংশ সমীকৃত করে

পাই, $\mathrm{a}=x^{3}-3 x y^{2}, \mathrm{~b}=3 x^{2} y-y^{3}$

এখन, $\mathrm{a}-i \mathrm{~b}=x^{3}-3 x y^{2}-i\left(3 x^{2} y-y^{3}\right)$

$=x^{3}-3 x y^{2}-3 x^{2} \cdot y+i y^{3}$

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

$=x^{3}-3 \cdot x^{2} \cdot i y+3 \cdot x \cdot(i y)^{2}-(i y)^{3}$

$=(x-i y)^{3}$

$\therefore \sqrt[3]{a+i b}=x-i y \quad$ (Showed)

এবং $\frac{a}{x}-\frac{b}{y}=\frac{x^{3}-3 x y^{2}}{x}-\frac{3 x^{2} y-y^{3}}{y}$

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

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

$\therefore-2\left(x^{2}+y^{2}\right)=\frac{a}{x}-\frac{b}{y} \quad$ (Showed)