If _{$z=x+iy,z^{1/3}=a-ib$} ,then _{$\frac{x}{a}-\frac{y}{b}=k\left(a^2-b^2\right)$} where k is equal to

  $z^{1/3}=a-ib\Rightarrow z={\left(a-ib\right)}^3$ 

$\therefore x+iy=a^3+i{b}^3-3i{a}^{2}b-3a{b}^2$ .

equate real parts on both sides     and  equate imaginary parts on both sides

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

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