Question

Factorise $2\sqrt{2}{a}^3+8b^3-27c^3+18\sqrt{2}abc$.

Answer 1

We transform this to the form
${\left(\sqrt{2}a\right)}^3+{\left(2b\right)}^3+{(-3c)}^3-3\left(\sqrt{2}a\right)\left(2b\right)(-3c)$
This is in the standard form $x^3+y^3+z^3-3xyz$, where $x=\sqrt{2}a,y=2b$ and $z=-3c$.
Using the identity
$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$
we obtain
$\sqrt{2}{a}^3+8b^3-27c^3+18\sqrt{2}abc=\left[\right(\sqrt{2}a)+2b-3c][{\left(\sqrt{2}a\right)}^2+{\left(2b\right)}^2+{(-3c)}^2-(\sqrt{2}a\left)\right(2b)-(2b\left)\right(-3c)-(-3c\left)\right(\sqrt{2}a\left)\right]$
This simplifies to
$2\sqrt{2}{a}^3+8b^3-27c^3+18\sqrt{2}abc=[\sqrt{2}a+2b-3c][{2a}^2+{4b}^2+9c^2-2\sqrt{2}ab+6bc+3\sqrt{2}a]$