If (x+iy)1/3=a+ib, then x/a+y/b=

x+iy=(a+ib)3 =a3+i3b3+3ai2b2+3ia2b =(a3−3ab2)+i(3a2b−b3) ⇒x=a3−3ab2, y=3a2b−b3 ⇒ x/a+y/b=(a2−3b2)+(3a2−b2)                             =4(a2−b2)