Factorize $x^4 + y^4 -x^2 y^2$ over $\mathbb C$ As in the title, I'd like to factorize $x^4 + y^4 - x^2y^2$ into irreducible factors over $\mathbb C$ (i.e. linear factors).
Attempts:
First I tried doing
$$x^4 + y^4 - x^2y^2 = (x^2 - y^2)^2 + x^2y^2 \\ = (x^2 - y^2) - (ixy)^2 \\ = (x^2 - y^2 - ixy)(x^2 - y^2 + ixy) $$
And I got stuck at this point.
Second (desperately), I expanded
$$(a_1 x + b_1y)(a_2 x + b_2 y)(a_3 x + b_3 y)(a_4 x + b_4 y)$$
and compared it to $x^4 + y^4 - x^2y^2$, but let's just say it didn't go smoothly.
I think there should be a nice way to do it. Anybody?
 A: Note that $x^2 - y^2 - ixy = y^2 \left( (\frac{x}{y})^2 - i(\frac{x}{y}) -1 \right)$, which is simply a quadratic in $(\frac{x}{y})^2$, multiplied by $y^2$. Factorise the quadratic and then multiply $y^2$ back in. Do the same with the other factor to finish.
A: A smart way to approach this is to solve the equation
$$x^2-y^2+ixy=0,$$ find the solutions $x_\pm(y)$ and then write
$$x^2-y^2+ixy=(x-x_+(y))(x_-(y)).$$
In this case we have that the solutions are
$$x_i(y)=\frac{-iy\pm\sqrt{(iy)^2+4y^2}}{2}=\frac{-iy\pm\sqrt{3}y}{2}=\frac{-1\pm\sqrt{3}}{2}\cdot y$$
thus the factorisation is
$$x^2-y^2+ixy=\left(x+\frac{-1+\sqrt{3}}{2}y\right)\left(x+\frac{-1-\sqrt{3}}{2}y\right).$$
The procedure is analogous for the second term
A: $$x^4-x^2y^2+y^4=(x^2+y^2)^2-3x^2y^2=(x^2-\sqrt3xy+y^2)(x^2+\sqrt3xy+y^2)=$$
$$=\left(\left(x-\frac{\sqrt3}{2}y\right)^2+\frac{y^2}{4}\right)\left(\left(x+\frac{\sqrt3}{2}y\right)^2+\frac{y^2}{4}\right)=$$
$$=\left(x-\frac{\sqrt3}{2}y+\frac{1}{2}yi\right)\left(x-\frac{\sqrt3}{2}y-\frac{1}{2}yi\right)\left(x+\frac{\sqrt3}{2}y+\frac{1}{2}yi\right)\left(x+\frac{\sqrt3}{2}y-\frac{1}{2}yi\right).$$
