Question

Factorise:
$a^3-b^3{+8c}^3+6abc$

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 $a=a,b=-b$ and $c=2c$, the equation $a^3-b^3+8c^3+6abc$ can be factorised as follows:

$a^3-b^3+8c^3+6abc=\left(a^3\right)+(-b{)}^3+(2c{)}^3-3\left(a\right)(-b)\left(2c\right)$

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

$=(a-b+2c)(a^2+b^2+4c^2+ab+2bc-2ca)$

Hence, $a^3-b^3+8c^3+6abc=(a-b+2c)(a^2+b^2+4c^2+ab+2bc-2ca)$