2017/006) if $x= a + b$ $y = aw+bw^2$ and $z= aw^2+ bw$ then show that $x^3+y^3 + z^3 = 3(a^3+b^3)$ and $x^2+y^2 + z^2 = 6ab$

We have $x+y+z = a(1+w+w^2) + b(1+w+w^2) = 0$
so x^3 + y^3+ z^3 = 3xyz = 3(a+b)(aw+bw^2)(aw^2+bw) = 3(a+b)(a^2 w^3 + abw^2 + abw^4 + b^2w^3)$= 3(a+b)(a^2 + ab(w^2+ w^4) + b^2) = 3(a+b)(a^2 + ab(w^2+w) + b^2)$
$= 3(a+b)(a^2 - ab + b^) = 3(a^3+b^3)$
Further
$x^2 = a^2 + b^2 + 2ab$
$y^2 = a^2w^2 + b^2w + 2ab$
$z^2 = a^2w + b^2w^2 + 2ab$
hence $z^2+y^2+z ^2 = a^2(1+w+w^2) + b^2(1+w+w^2) + 6ab = 6ab$