Question

Factorise $a^3+8b^3+c^3-6abc$ using the identity $a^3+b^3+c^3-3abc=\frac{1}{2}(a+b+c)[{(a-b)}^2+{(b-c)}^2+{(c-a)}^2]$

Answer 1

Using $a^3+b^3+c^3-3abc=\frac{1}{2}(a+b+c)[{(a-b)}^2+{(b-c)}^2+{(c-a)}^2]$

We can rewrite, $a^3+8b^3+c^3-6abc$ as,

$=\frac{1}{2}(a+2b+c)[{(a-2b)}^2+{(2b-c)}^2+{(c-a)}^2]$

Hence, the answer is $\frac{1}{2}(a+2b+c)[{(a-2b)}^2+{(2b-c)}^2+{(c-a)}^2].$