Question

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

The given identity is $a^3+b^3+c^3-3abc=\frac{1}{2}(a+b+c)\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]$

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

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

$=\frac{1}{2}(1+b+2c)(1+b^2-2b+b^2+4c^2-4bc+4c^2+1-4c)=\frac{1}{2}(1+b+2c)(2+2b^2+8c^2-2b-4bc-4c)$

Hence, $1^3+b^3+8c^3-6bc=\frac{1}{2}(1+b+2c)(2+2b^2+8c^2-2b-4bc-4c)$