<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] $(a^3+b^3+c^3) = (a+b+c)(a^2 +b^2+c^2-ab-bc-ac)$

$1+\omega + \omega^2 = 0 \implies \omega+\omega^2 = -1$

$= (a+b+c)(a^2 + b^2 + c^2 +ab(\omega+\omega^2)+bc(\omega+\omega^2)+ac(\omega+\omega^2)) = (a+b+c)(a(a+b\omega^2+c\omega)+b\omega(a+b\omega^2+ c\omega)+ c\omega^2 (a+ \dfrac b\omega+c\omega)$
Now, note that $\dfrac{1}{\omega}= \omega^2$
And after substituting this we get the desired identity. 
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