If a+b+c=0, and w is a complex root of cube roots of unity, then can...

Let ${\displaystyle X=(a+bw+c{w}^2)\text{and}Y=(a+b{w}^2+cw)}$.
The key idea is to note that
${\displaystyle X^3+Y^3=(X+Y)(X+wY)(X+w^{2}Y)}$ for all X,Y
This, in turn, follows from using the cube roots of unit ${\displaystyle (1,w,w^2)}$ to factor ${\displaystyle t^3-1}$:
${\displaystyle t^3-1=(t-1)(t-w)(t-w^2)}$
And then substituting ${\displaystyle t=\frac{x}{-y}}$ on both sides.
Now, using ${\displaystyle 1+w+w^2=0}$ (sum of roots in ${\displaystyle t^3-1=0}$ by Vietta) repeatdly, observe that:
1) ${\displaystyle X+Y=2a-b-c=3a}$
2) ${\displaystyle X+wY=a+bw+c{w}^2+w(a+b{w}^2+cw)=a(1+w)+b(w+1)+2c{w}^2=3c{w}^2}$
3) ${\displaystyle X+w^{2}Y=a+bw+c{w}^2+w^2(a+b{w}^2+cw)=a(1+w^2)+2bw+c(w^2+1)=3bw}$
Multiplying out (i), (ii) and (iii), we get:
${\displaystyle X^3+Y^3=(X+Y)(X+wY)(X+w^{2}Y)=27abc{w}^3=27abc}$

Let ${\displaystyle U=a+b\omega +c{w}^2\text{and}V=a+b{\omega }^2+c\omega }$
${\displaystyle UV=a^2+(ab+ac)\omega +(ab+bc+ca){\omega }^2+b^2{\omega }^3+c^2{\omega }^3+bc{\omega }^4}$
${\displaystyle =a^2+b^2+c^2+(ab+bc+ca)(\omega +{\omega }^2)\text{since}{\omega }^4=\omega }$
${\displaystyle =a^2+b^2+c^2-(ab+bc+ca)}$
${\displaystyle ={(a+b+c)}^2-3(ab+bc+ca)}$
${\displaystyle =-3(ab+bc+ca)\text{since}a+b+c=0}$
${\displaystyle =-3(a(b+c)+bc)}$
${\displaystyle =-3a^2-3bc\text{since}b+c=-a}$
${\displaystyle U^3+V^3={(U+V)}^3-3UV(U+V)}$
${\displaystyle ={\left(3a\right)}^3-3(-3a^2-3bc)\cdot 3a}$
${\displaystyle =27a^3-27a^3+27abc}$
${\displaystyle =27abc}$