How to arrive from $x^2y^2－x^2－y^2－6xy＋4$ to $(xy＋x＋y－2）(xy－x－y－2）$?

Question

How do we arrive from

$x^2y^2－x^2－y^2－6xy＋4$

to

$(xy＋x＋y－2）(xy－x－y－2）$

?

According to my book these two are equal, but I can't understand how to transform one to another.

Answer 1

$$x^2y^2-x^2-y^2\color{red}{-6xy}+4=x^2y^2\color{red}{-4xy}+4-x^2\color{red}{-2xy}-y^2\\=(xy-2)^2-(x+y)^2=(xy+x+y-2)(xy-x-y-2)$$

Answer 2

Start with $$(xy+x+y-2)(xy-x-y-2)=$$

$$ [(xy-2)+(x+y)][(xy-2)-(x+y)]=$$

$$(xy-2)^2-(x+y)^2$$

You can take it from here.

Answer 3

Use that $x^2y^2-4xy+4=(xy-2)^{2}$and $(x+y)^{2}=x^2+2xy+y^2.$ So, you have

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

Use the fact that $a^{2}-b^{2}=(a+b)(a-b)$, and you are done

Answer 4

Just splitting the $6xy$ into two parts: \begin{align} x^2y^2－x^2－y^2－6xy＋4&=x^2y^2\color{red}{-x^2-y^2-2xy}-4xy+4\\ &=(xy-2)^2-\color{red}{(x+y)^2}=\dotsb \end{align}

Answer 5

Notice we have both $x^2$ and $y^2$ with the same coefficient and something of the form $axy$ with a coefficient of the same sign. We can pull out an $(x+y)^2$ from this to get $$x^2y^2-4xy+4-(x+y)^2$$ Notice now that $x^2y^2-4xy+4$ is $(xy-2)^2$. Then we have $$(xy-2)^2-(x+y)^2$$ This is a difference of two squares, so we have $$(xy-2+x+y)(xy-2-(x+y))$$