Question

Resolve $a^3-b^3+c^3+3abc$ in to factors.

Answer 1

We write this in the form.
$a^3+{(-b)}^3+c^3-3a(-b)c$.
This in the familiar form, where $b$ is replaced by $-b$. Thus
$a^3+{(-b)}^3+c^3-3a(-b)c=[a+(-b)+c][a^2+{(-b)}^2+c^2-a(-b)-(-b)c-ca]$
Simplification gives
$a^3-b^3+c^3+3abc=(a-b+c)(a^2+b^2+c^2+ab+bc-ca)$