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.
 A: 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.
A: 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
$$(x-jy)(x-j^2y)\quad\text{where } j\; \text{and }j^2 \;\text{are the non-real cubic roots of unity.}$$
A: 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
A: 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
A: I am sorry, there is a slight mistake, I forgot factoring 2. Thus $x^4 + y^4 + (x+y)^4 = 2(x^2 + xy + y^2)^2$
A: $$(x+y)^4+x^4+y^4=(x^2+y^2+2xy)^2+(x^2+y^2)^2-2(xy)^2=\\
[(x^2+y^2+2xy)^2-(xy)^2]+[(x^2+y^2)^2-(xy)^2]=\\
(x^2+y^2+xy)(x^2+y^2+3xy)+(x^2+y^2+xy)(x^2+y^2-xy)=\\
(x^2+y^2+xy)(2x^2+2y^2+2xy)=2(x^2+y^2+xy)^2$$
A: $$x^4 + y^4 + (x + y)^4
=2 x^4 + 4 x^3 y + 6 x^2 y^2 + 4 x y^3 + 2 y^4$$
$$=x (x (x (2 x + 4 y) + 6 y^2) + 4 y^3) + 2 y^4
=2 x^4 + y (4 x^3 + y (6 x^2 + y (4 x + 2 y)))$$
$$=4 y^2 (x^2 + x y) + 2 (x^2 + x y)^2 + 2 y^4
=2 (x^4 + 2 x^3 y + 3 x^2 y^2 + 2 x y^3 + y^4)$$
$$=2 (x^2 + x y + y^2)^2$$
