If z=x+iy,z1/3=a−ib ,then xa−yb=k(a2−b2) where k is equal to

z1/3=a−ib⇒z=(a−ib)3 ∴x+iy=a3+ib3−3ia2b−3ab2 .equate real parts on both sides     and  equate imaginary parts on both sides then x=a3−3ab2⇒xa=a2−3b2 ,    y=b3−3a2b⇒yb=b2−3a2 So,xa−yb=4(a2−b2)