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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?

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    Have you seen this? – rtybase Nov 01 '20 at 12:26
  • Hint: As you might know from your inequalities training, $a^2 + b^2 + c^2 - bc - ca - ab = \dfrac12\left(x^2 + y^2 + \left(x-y\right)^2\right)$ where $x = b-c$ and $y = c-a$. So now you need to factor a homogeneous polynomial in two variables $x$ and $y$. This is tantamount to factoring a (non-homogeneous) polynomial in one variable. – darij grinberg Nov 01 '20 at 12:28
  • Here is another one, pages 3-4 in preview mode. – rtybase Nov 01 '20 at 12:29

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

Zerox
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