If ${(x+iy)}^{\frac{1}{3}}=a+ib$, then $\frac{x}{a}+\frac{y}{b}=$

${(x+iy)}^{\frac{1}{3}}=a+ib$

cubing on both sides

$$\begin{gathered} x+iy={(a+ib)}^3 \\ x+iy=a^3+3a^2\left(ib\right)+3a{\left(ib\right)}^2+{\left(ib\right)}^3 \\ x+iy=a^3-3a{b}^2+i(3a^{2}b-b^3) \end{gathered}$$

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

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

$\frac{x}{a}+\frac{y}{a}=a^2-3b^2+3a^2-b^2=4a^2-4b^2$

$\frac{x}{a}+\frac{y}{b}=4(a^2-b^2)$