Question

Resolve into factors: $x^4-6x^2{y}^2+y^4$

Answer 1

In the given expression $x^4+y^4-6x^2{y}^2$, add and subtract $2x^2{y}^2$ to make it a perfect square as shown below:

$x^4+y^4-6x^2{y}^2=(x^4+y^4+2x^2{y}^2)-6x^2{y}^2-2x^2{y}^2=(x^2+y^2{)}^2-8x^2{y}^2$

(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 $(x^2+y^2{)}^2-8x^2{y}^2$can be factorised as follows:

$(x^2+y^2{)}^2-8x^2{y}^2=(x^2+y^2{)}^2-(\sqrt{8}xy{)}^2=(x^2+y^2+\sqrt{8}xy\left)\right(x^2+y^2-\sqrt{8}xy)$

Hence, $x^4+y^4-6x^2{y}^2=(x^2+y^2+\sqrt{8}xy)(x^2+y^2-\sqrt{8}xy)$