2016/051) If $1, w, w^2$ are cube roots of unity and $a+b+c=0$ then prove that $(a+bw+cw^2)^3+(a+bw^2+cw)^3=27abc$

let $X = a+bw+cw^2$
$Y = a+bw^2+cw$
$X+Y = 2a + b(w+w^2) + c(w^2+w) = 2a - b -c = 3a - (a+b+c) = 3a$
let $Z = -3a$
so $X+Y+Z=0$
so $X^3 + Y^3 + Z^3 = 3XYZ = -9a(a+bw+cw^2)(a+bw^2+cw)$
           $ = -9a(a^2 + b^2 + c^2 - ab - ac - bc)$
or $X^3 + Y^3 + (-3a)^3 = -9a((b+c)^2 + b^2 + c^2 - a(b +c) - bc)$
or $X^3 + Y^3 + (-3a)^3 = -9a((b+c)^2 + (b+c)^2 - 2bc + (b +c)^3 - bc)$
or $X^3 + Y^3 = 27a^3 -9a(3(b+c)^2 + 3 bc)$
$ = 27a(a^2 -  (b+c)^2  + bc)$
$= 27a((b+c)^2 - (b+c)^2 + bc) = 27abc$
hence  $(a+bw+cw^2)^3+(a+bw^2+cw)^3=27abc$