Question

Factorise: $64a^4+1$

Answer 1

In the given equation $64a^4+1$, add and subtract $16a^2$ to make it a perfect square as shown below:

$64a^4+1=(64a^4+1+16a^2)-16a^2=\left[\right(8a^2{)}^2+(1{)}^2+2(8a^2\left)\right(1\left)\right]-16a^2=(8a^2+1{)}^2-16a^2$

(using the identity $(a+b{)}^2=a^2+b^2+2ab$

We also know the identity $a^2-b^2=(a+b)(a-b)$, therefore,

Using the above identity, the equation $(8a^2+1{)}^2-16a^2$can be factorised as follows:

$(8a^2+1{)}^2-16a^2=(8a^2+1{)}^2-(4a{)}^2=(8a^2+1+4a\left)\right(8a^2+1-4a)$

Hence, $64a^4+1=(8a^2+1+4a)(8a^2+1-4a)$