Question

Resolve into factors: $81a^4+9a^2{b}^2+b^4$

Answer 1

We know the identity $a^4+b^4+a^2{b}^2=(a^2+b^2+ab)(a^2+b^2-ab)$

Using the above identity, the equation $81a^4+b^4+9a^2{b}^2$ can be factorised as follows:

$81a^4+b^4+9a^2{b}^2=(3a{)}^4+(b{)}^4+\left(3a{)}^2\right(b{)}^2=\left[\right(3a{)}^2+(b{)}^2+(3a\left)\right(b\left)\right]\left[\right(3a{)}^2+(b{)}^2+(3a\left)\right(b\left)\right]$

$=(9a^2+b^2+3ab)(9a^2+b^2-3ab)$

Hence, $81a^4+9a^2{b}^2+b^4=(9a^2+b^2+3ab)(9a^2+b^2-3ab)$