<s>[INST] How to show that $A^3+B^3+C^3 - 3ABC = (A+B+C)(A+B\omega+C\omega^2)(A+B\omega^2+C\omega)$ indirectly? I found this amazingly beautiful identity here. How to prove that $A^3+B^3+C^3 - 3ABC = (A+B+C)(A+B\omega+C\omega^2)(A+B\omega^2+C\omega)$ without directly multiplying the factors? (I've already verified it that way). Moreover, how could someone possibly find such a factorization using complex numbers? Is it possible to find such a factorization because $A^3+B^3+C^3 - 3ABC$ is a symmetric polynomial in $A,B,C$?
 [/INST] Expanding on Potato's answer:
First lets make $a^3 + b^3 + c^3 -3abc\ $ a polynomial in $a$ so we get $a^3 - a3bc + b^3 + c^3$. This has $3$ roots so lets express it as $(a+p)(a+q)(a+r)$. But also know: $$a^3 + b^3 + c^3 -3abc = (a+b+c)(a^2 + b^2 + c^2 -ab -bc - ac)$$
Therefore lets say $p=b+c$. So our polynomial is now: $(a+q)(a+r)(a+b+c)$. Expanding this expression and equating the co-efficients of $a$ we get the following equations:
$$r+q + b+c = 0 $$
$$qr+br+bq+cr+cq = -3bc$$
$$bqr + cqr = b^3 + c^3$$
Doing sum of roots on the third equation we can get:
$$qr = b^2 - bc +c^2$$
subbing the this equation into the second equation we get:
$$(b+c)^2 + br+bq+cr+cq = 0$$
$$(b+c)^2 = -(b+c)(q+r)$$
$$b+c = -(q+r)$$
Now, since $b+c$ is real and lets assume $q$ and $r$ are complex, this means $r$ is the conjugate of $q$. So let $q=x+yi$. Therefore we get:
$$-2x = b+c$$
$$qr = |q|^2 = x^2+y^2 = b^2+c^2-bc$$
So we have $x = -(b+c)/2$ and substituting this in the above equation we get $y = \frac{\sqrt{3}}{2}(b-c)$.
This means:
$$q= -\frac{b+c}{2} + \frac{\sqrt{3}}{2}(b-c)i$$
$$ = -\frac{b}{2} + \frac{\sqrt{3}}{2}bi + -\frac{c}{2} - \frac{\sqrt{3}}{2}ci$$
$$ = bw^2 + cw$$
where w is the complex cubic root of unity. Hence this means that:
$$ r = cw^2 + bw$$
Therefore:
$$a^3 + b^3 + c^3 -3abc = (a+b+c)(a+  bw + cw^2)(a+bw^2 + cw)$$
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