(b) ${(x + iy)^{1/3}} = a + ib$==>$(x + iy) = {(a + ib)^3}$
$ = {a^3} + 3{a^2}.ib + 3a.{(ib)^2} + {(ib)^3}$
$ = {a^3} – 3a{b^2} + i(3{a^2}b – {b^3})$
Equating real and imaginary parts, we get
$\frac{x}{a} = {a^2} – 3{b^2}$and $\frac{y}{b} = 3{a^2} – {b^2}$
$\therefore $ $\frac{x}{a} + \frac{y}{b} = 4({a^2} – {b^2})$