Question

Factorise:
$1+b^3+8c^3-6bc$

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 $a=1,b=b$ and $c=2c$, the equation $1+b^3+8c^3-6bc$ can be factorised as follows:

$1+b^3+8c^3-6bc=(1{)}^3+(b{)}^3+(2c{)}^3-3(1\left)\right(b\left)\right(2c)$

$=(1+b+2c)\left[\right(1{)}^2+(b{)}^2+(2c{)}^2-(1\times b)-(b\times 2c)-(2c\times 1)]=(1+b+2c\left)\right(1+b^2+4c^2-b-2bc-2c)$

Hence, $a+b^3+8c^3-6bc=(1+b+2c)(1+b^2+4c^2-b-2bc-2c)$