<s>[INST] How would I factor $a^3+b^3+c^3-6abc$ How would I factor the polynomial $a^3+b^3+c^3-6abc$? The values are homogenous, so so must be the factors. I don't know where to go from there.
 [/INST] Try solving the cubic for $c=x+y$ so that $c^3=(x^3+y^3)+3xy(x+y)=(x^3+y^3)+3xyc$
Whence $x^3+y^3=-a^3-b^3$ and $xy=2ab$ so that $x^3y^3=8a^3b^3$ and $x^3, y^3$ are roots of the quadratic $$z^2+(a^3+b^3)z+8a^3b^3$$ so that $$x^3,y^3=\frac {-a^3-b^3\pm  \sqrt {a^6+b^6-30a^3b^3}}2$$
And use this to build your factors $(c-x-y)(c-\omega x-\omega^2 y)(c-\omega^2 x - \omega y)$ where $\omega$ is a non-trivial cube root of $1$.
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