<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] HINT:
If $A+Bw+Cw^2=0$ where $w$ is one of the three cube roots of unity 
$\implies -A=Bw+Cw^2$
Cubing we get, $(-A)^3=(Bw+Cw^2)^3$
$\implies  -A^3=B^3w^3+C^3w^6+3\cdot Bw\cdot Cw^2(Bw+Cw^2)=B^3+C^3+3BC(-A)$
$\implies A+Bw+Cw^2$ is a factor of $A^3+B^3+C^3-3ABC$
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