Factorisation of $a^3-b^3+8c^3+6abc$

Question

I came across this question which states " Factorise this : $a^3-b^3+8c^3-6abc$". This confused me and I'm not even sure if this is factorable however It'll be cool to see how it's factorised, if it can be.

Answer 1

Sonnhard got the right idea, but made one small typo. The correct answer is $$a^3-b^3+8c^3+6abc=(2c+a-b)\left(a^2+b^2+4c^2+ab-2ac+2bc\right).$$ The expression $a^3-b^3+8c^3-6abc$ is irreducible over $\mathbb Z$.

Answer 2

HINT: One factor is $$2c+a-b$$ Multiply out $$(2c+a-b)(a^2+ab-2ac+b^2+2bc+4c^2)$$

Answer 3

Use the standard identity: \begin{align}x^3+y^3+z^3-3xyz&=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\\&=\tfrac12(x+y+z)\bigl((x-y)^2+(y-z)^2+(z-x)^2\bigr)\end{align}

Answer 4

Using identity $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$
Which gives $$a^3-b^3+8c^3-6abc = (a-b+2c)(a^2+b^2+4c^2+ab+2bc-2ca)$$