If ${(x+iy)}^{\frac{1}{3}}=a+ib$, where $i=\sqrt{-1}$, then $(\frac{x}{a}+\frac{y}{b})$ is equal to

$$\begin{gathered} {(x+iy)}^{\frac{1}{3}}=a+ib \\ \Rightarrow (x+iy)={(a+ib)}^3 \Rightarrow x+iy=a^3-i{b}^3+3a^{2}bi-3a{b}^2 \\ \Rightarrow x=a^3-3a{b}^2 \& y=-b^3+3a^{2}b \\ \Rightarrow x=a[a^2-3b^2] \& y=b[-b^2+3a^2] \\ \Rightarrow \frac{x}{a}=a^2-3b^2; \frac{y}{b}=-b^2+3a^2 \\ \Rightarrow \frac{x}{a}+\frac{y}{b}=a^2-3b^2-b^2+3a^2 \\ \Rightarrow \frac{x}{a}+\frac{y}{b}=4a^2-4b^2=4(a^2-b^2) \\ \Rightarrow \frac{x}{a}+\frac{y}{b}=4(a^2-b^2) \end{gathered}$$