Factoring $x^4-11x^2y^2+y^4$ I am brushing up on my precalculus and was wondering how to factor the expression
$$
x^4-11x^2y^2+y^4
$$
Thanks for any help!

For emphasis, if you have a quadratic or quadratic form, $a x^2 + bx + c$ or $a x^2 + bxy + c y^2,$it factors over the integers if and only if the discriminant from the Quadratic Formula, $b^2 - 4 a c,$ is an integer squared. 
Things are rather different for quartics with only EVEN exponents, either 
$f^2 x^4 + g x^2 + h^2$ or $f^2 x^4 + g x^2 y^2 + h^2 y^4$ to keep it easy. No worry about the discriminant, but the only thing that can work (once no rational roots) is, where $fh$ could be positive or negative,
$$ (fx^2 + \gamma xy + h y^2 )  (fx^2 - \gamma xy + h y^2 )  = f^2 x^4 + g x^2 y^2 + h^2 y^4.  $$
Here we just need to be able to solve for $\gamma$ in
$$ 2fh - \gamma^2 = g,  $$
or
$$ 2fh - g = \gamma^2.  $$
To repeat, we can negate $fh$ at need, useful if $g$ is negative. We see this in the original question and the answer by 6005, where we can choose either $ 2 + 11 = 13$ or $-2 + 11 = 9.$
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$$ (x^2 + 2 xy + 2 y^2)(x^2 - 2xy + 2 y^2) = x^4 + 4 y^4, $$
$$ (x^2 +  xy +  y^2)(x^2 - xy +  y^2) = x^4 + x^2 y^2 + y^4. $$
The latter gives a real analytic alternative to the superellipse.  

