$x=a+b$
$\begin{aligned} & y=a \alpha+b \beta, z=a \beta+b \alpha \\ & \alpha=\omega \text { and } \beta=\omega^2 \\ & \because x+y+z=(a+a \alpha+a \beta)+(b+b \beta+b \alpha) \\ & =a(1+\alpha+\beta)+b(1+\beta+\alpha) \\ & =a\left(1+\omega+\omega^2\right)+b\left(1+\omega+\omega^2\right)=0 \\ & \therefore x^3+y^3+z^3=3 x y z \\ & =3(a+b)\left(a \omega+b \omega^2\right)\left(a \omega^2+b \omega\right)\end{aligned}$
$\begin{array}{r}=3\left(a^3 \omega^3+a^2 b \omega^2+a^2 b \omega^4+a b^2 \omega^3+b a^2 \omega^3+b^2 a \omega^2+\right. \\ a b^2 \omega^2+b^3 \omega^3\end{array}$
$\begin{aligned} & =3\left(a^3+b^3+a^2 b \omega^2+a^2 b \omega+a b^2+a^2 b+a b^2 \omega^2+a b^2 \omega^2\right) \\ & =3\left[a^3+b^3+a b^2\left(1+\omega+\omega^2\right)+a^2 b\left(\omega^2+\omega+1\right)\right] \\ & =3\left(a^3+b^3\right)\end{aligned}$