book shows answer as x^4+4x^2y^2+4y^4-4x^2y^2=(x^2+2y^2)^2-(2xy)^2=(x^2+2y^2-2xy)(x^2+2y^2+2xy) What is happening here and what steps am i missing???
$\displaystyle x^4+4x^2y^2+4y^4-4x^2y^2$
$\displaystyle {\color{red} (x^4+4x^2y^2+4y^4)} -4x^2y^2$
note these terms form a perfect square ...
$\displaystyle {\color{red} (x^2+2y^2)^2} - (2xy)^2$
now you have the difference of two squares ... $\displaystyle a^2-b^2 = (a-b)(a+b)$
$\displaystyle [(x^2+2y^2)-2xy] \cdot [(x^2+2y^2)+2xy]$
What they have done is add and subtract $\displaystyle 4x^2y$ which, of course, doesn't change the actual value. But that does change the form so that it we can think of it as $\displaystyle x^4+4x^2y^2+4y^4-4x^2y^2= (x^2+ 2y^2)^2- 4x^2y^2$ which is a "difference" of two squares. You probably know that $\displaystyle a^2- b^2= (a- b)(a+ b)$ so that last is $\displaystyle ([x^2+ 2y^2]+ 2xy)([x^2+ 2y^2]- 2xy)= (x^2+ 2xy+ 2y^2)(x^2- 2xy+ 2y^2)$.
another example ... let's say you wish to factor $\displaystyle y = x^4 + 1$
$\displaystyle x^4 + 2x^2 + 1 - 2x^2$
$\displaystyle (x^2 + 1)^2 - (\sqrt{2} x)^2$
$\displaystyle [(x^2+1) - \sqrt{2} x][(x^2 + 1) + \sqrt{2} x]$
$\displaystyle (x^2 - \sqrt{2} x + 1)(x^2 + \sqrt{2} x + 1)$