Proving inequality using root of unity Let $\omega$ be a complex cube root of unity. It can be shown that if $a,b,c \in \mathbb{R}$, then $$(a+b+c)(a+b\omega+c\omega^2)(a+b \omega^2 + c \omega)= a^3+b^3+c^3-3abc$$
I was wondering if this could be extended to prove the AM/GM inequality for $n=3$. All that needs to be done is to prove that the left expression is non-negative, but I cannot find a way.
 A: Proof of AM-GM:
Using $\omega^3=1, \omega^2+\omega+1=0$, We know that $(a+b\omega+c\omega^2)(a+b \omega^2 + c \omega)=a^2+b^2+c^2-ab-bc-ca$, then
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\ge 0$
If $a,b,c>0$, then  $a+b+c>0$ and $\frac{a^2+b^2}{2}\ge ab$ etc give $a^2+b^2+c^2\ge ab+bc+ca$
Hence $\frac{a^3+b^3+c^3}{3}\ge abc \implies \frac{x+y+z}{3} \ge (xyx)^{1/3}.$
A: 
Leading the proof of $n=3$-AM-GM Ineq. from:
$$(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)=a^3+b^3+c^3-3abc.$$

Enough to show that: $(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega) \ge 0.$
Let's make all of the conditions of equality the same in the proof of the above.
\begin{align}
\newcommand{w}{\omega}
& \text{WLOG in RHS, let } a\ge b\ge c.\\
&(a+b+c)(a+b\w+c\w^2)(a+b\w^2+c\w) \\
=&(a+b+c)\Big((a-c)+(b-c)\w\color{red}{+(c\w^2+c\w+c)}\Big)\Big((a-b)+(c-b)\w\color{red}{+(b\w^2+b\w+b)}\Big) \\
=&(a+b+c)\Big((a-c)+(b-c)\w\Big)\Big((a-b)+(c-b)\w\Big) \\
=&(a+b+c)\Big((a-b)(a-c)+(a-b)(b-c)\w+(a-c)(c-b)\w-(b-c)^2\w^2\Big) \\
=&(a+b+c)\Big((a-b)(a-c)+(a-b+c-a)(b-c)\w+(b-c)^2(\w+1)\Big) \\
=&(a+b+c)\Big((a-b)(a-c)-(b-c)^2\w+(b-c)^2\w+(b-c)^2\Big) \\
=&(a+b+c)\Big((a-b)(a-c)+(b-c)^2\Big) \\
\ge&(a+b+c)\Big((a-a)(a-a)+(b-b)^2 \Big) \\
= & 0.
\end{align}
Condition of Equality: $b=a, c=a, c=b \Rightarrow a=b=c.$
