Question

Factorise:
$125a^3+b^3+64c^3-60abc$

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 $a=5a,b=b$ and $c=4c$, the equation $125a^3+b^3+64c^3-60abc$ can be factorised as follows:

$125a^3+b^3+64c^3-60abc=\left(5a^3\right)+(b{)}^3+(4c{)}^3-3\left(5a\right)\left(b\right)\left(4c\right)$

$=(5a+b+4c)\left[\right(5a{)}^2+(b{)}^2+(4c{)}^2-(5a\times b)-(b\times 4c)-(4c\times 5a)]$

$=(5a+b+4c)(25a^2+b^2+16c^2-5ab-4bc-20ca)$

Hence, $125a^3+b^3+64c^3-60abc=(5a+b+4c)(25a^2+b^2+16c^2-5ab-4bc-20ca)$