How would I factor $a^3+b^3+c^3-6abc$

Question

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.

Answer 1

$$a^3+b^3+c^3-6abc= (a+b)^3-3ab(a+b)+c^3-6abc \\(a+b+c)^3-3(a+b)c(a+b+c) -3ab(a+b)-6abc \\=(a+b+c)^3 -3(a+b)c(a+b+c)-3ab(a+b)-(3+3)abc \\=(a+b+c)^3-3c((a+b)(a+b+c)) -3ab(a+b+c) $$ now factor $$ (a+b+c) $$ and go on $$(a+b+c)((a+b+c)^2 -3c(a+b) -3ab) ) \\=(a+b+c)(a^2+b^2+c^2+2ab+2ac+2bc-3ab-3bc-3ac) \\=(a+b+c)(a^2+b^2+c^2-ab-ac-bc) $$

Answer 2

Doesn't work. Over $\mathbb C,$ the closest you can get, with $\beta^3 = 1$ and $\gamma^3 = 1,$ is $$ (A + \beta B + \gamma C) (A^2 + \beta^2 B^2 + \gamma^2 C^2 - \beta \gamma BC - \gamma CA - \beta AB) = A^3 + B^3 + C^3 - 3 \beta \gamma ABC $$

Answer 3

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 $1$.