Question

Factorise:
$2\sqrt{2}{a}^3+16\sqrt{2}{b}^3+c^3-12abc$

Answer 1

We know the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

Using the above identity taking $a=\sqrt{2}a$, $b=2\sqrt{2}b$ and $c=c$, the equation can be factorised as follows:

$2\sqrt{2}{a}^3+16\sqrt{2}{b}^3+c^3-12abc=(\sqrt{2}a{)}^3+(2\sqrt{2}b{)}^3+(c{)}^3-3(\sqrt{2}a\left)\right(2\sqrt{2}b\left)\right(c)$

$=(\sqrt{2}a+2\sqrt{2}b+c)\left[\right(\sqrt{2}a{)}^2+(2\sqrt{2}b{)}^2+(c{)}^2-(\sqrt{2}a\times 2\sqrt{2}b)-(2\sqrt{2}b\times c)-(c\times \sqrt{2}a)]$

$=(\sqrt{2}a+2\sqrt{2}b+c)(2a^2+8b^2+c^2-4ab+2\sqrt{2}bc-\sqrt{2}ca)$

Hence, $2\sqrt{2}{a}^3+16\sqrt{2}{b}^3+c^3-12abc=(\sqrt{2}a+2\sqrt{2}b+c)(2a^2+8b^2+c^2-4ab+2\sqrt{2}bc-\sqrt{2}ca)$