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?

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 $A,B,C$?