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.

Note that $$(x+y)^4 + x^4 + y^4 = y^4 ((x/y+1)^4 + (x/y)^4 + 1)$$ and see if you can factor $((t+1)^4 + t^4 + 1$. There is a quadratic factor.

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 $$(x-jy)(x-j^2y)\quad\text{where } j\; \text{and }j^2 \;\text{are the non-real cubic roots of unity.}$$

Beyond what I have posted, it cannot be factored in the real field. If you want to factor it in the complex field, you have to learn how to solve a quartic equation because you essentially need to know that roots of $(t+1)^4 + t^4 + 1 = 2t^4 + 4t^3 + 6t^2 + 4t + 2 = 0$, it can be simplified to $t^4 + 2t^3 + 3t^2 + 2t + 1 = 0$.Lucky enough, this is equal to $(t^2 + t + 1)^2$

Thus $x^4 + y^4 + (x + y)^4 = 2(x^2 + xy + y^2)^2$. No further factorization is available in the real field...

Best wishes

I know that using complex algebra, we can factor $x^4 + y^4 = (x^2 + y^2 + \sqrt{2}xy)(x^2 + y^2 - \sqrt{2}xy)$. I have no idea how to proceed forward... Good luck