Factorise:
$\displaystyle (xy+1)^4 - 4xy(xy+1)^2 - (x^2 - y^2)^2$
Thank you.
$\displaystyle (xy+1)^4 - 4xy(xy+1)^2 - (x^2 - y^2)^2$
As you see, $\displaystyle (xy+1)^2$ comes twice, so, Complete the square
$\displaystyle =[(xy+1)^2]^2-2(2xy)(xy+1)^2+(2xy)^2-(2xy)^2-(x^2-y^2)^2$
$\displaystyle =[(xy+1)^2-(2xy)]^2-(2xy)^2-(x^2-y^2)^2$
$\displaystyle =[x^2y^2+1+2xy-2xy]^2-4x^2y^2-[(x^2)^2+(y^2)^2-2x^2y^2]$
$\displaystyle =(x^2y^2+1)^2-[4x^2y^2+(x^2)^2+(y^2)^2-2x^2y^2]$
$\displaystyle =(x^2y^2+1)^2-[(x^2)^2+(y^2)^2+2x^2y^2]$
$\displaystyle =(x^2y^2+1)^2-(x^2+y^2)^2$
$\displaystyle =(x^2y^2+1+x^2+y^2)(x^2y^2+1-x^2-y^2)$