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

Solution :
$$\eqalign{ & {z^{\frac{1}{3}}} = a - ib \cr & \Rightarrow z = {\left( {a - ib} \right)^3} \cr & \therefore x + iy = {a^3} + i{b^3} - 3i{a^2}b - 3a{b^2}. \cr & {\text{Then, }}x = {a^3} - 3a{b^2} \cr & \Rightarrow \frac{x}{a} = {a^2} - 3{b^2} \cr & y = {b^3} - 3{a^2}b \cr & \Rightarrow \frac{y}{b} = {b^2} - 3{a^2} \cr & {\text{So, }}\frac{x}{a} - \frac{y}{b} = 4\left( {{a^2} - {b^2}} \right) \cr} $$