Completely factoring $a^3 + b^3 + c^3 - 3abc$?

Question

How do I completely factor $a^3 + b^3 + c^3 - 3abc$ over the complex numbers? The first thing I did is factor this into $(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac)$. But I am not sure how to factor $a^2 + b^2 + c^2 - ab - bc - ac$ over the complex numbers. Can anyone help me?

Answer 1

$a^3+b^3+c^3-3abc=(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)$, where $\omega$ is the root of $x^2+x+1=0$.