Factor $(x+y)^4+x^4+y^4$ Title says it all, I just want to know how to factor $(x+y)^4+x^4+y^4$. I only know that it's possible to factor, but got no idea how to do it. If it were a single-variable polynomial I could try to find rational roots or something, but I'm lost with this one.

This is a symmetric polynomial in $x$ and $y$, hence it can be expressed, by Newton's theorem, as a polynomial in $s=x+y$ and $p=xy$.
Indeed $x^2+y^2=(x+y)^2-2xy=s^2-2p$, whence
$$x^4+y^4=(x^2+y^2)^2-2x^2y^2=(s^2-2p)^2-2p^2=s^4-4ps^2+2p^2.$$
Therefore
$$(x+y)^4+x^4+y^4=2s^4-4ps^2+2p^2=2(s^2-p)^2=2(x^2+xy+y^2)^2.$$
$x^2+xy+y^2$ is irreducible in $\mathbf R[x]$, but  factors in  $\mathbf C[x]$ as
$$(x-jy)(x-j^2y)\quad\text{where } j\; \text{and }j^2 \;\text{are the non-real cubic roots of unity.}$$
