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 $A,B,C$?

We know $(A+B)\mid(A^n+B^n)$ for $n$ odd. What about with three terms? Compute
$$\mod A+B+C:\quad A^3+B^3+C^3\equiv-(B+C)^3+B^3+C^3\equiv-3BC(B+C)\equiv 3ABC.$$
So $(A+B+C)\mid(A^3+B^3+C^3-3ABC)=f(A,B,C)$. Further
$$f(A,B,C)=f(A,\omega B,\bar{\omega}C)=f(A,\bar{\omega}B,\omega C)=\rm etc.$$
by inspection so both $A+\omega B+\bar{\omega}C$ and $A+\bar{\omega}B+\omega C$ are also factors.
This argument exploits divisibility properties and inherent symmetry. It is generalized by the first proof (using matrix operations) mentioned in my other answer to compute $\Phi(G)$ for $G$ abelian.
