Factorise $x^4+y^4+(x+y)^4$ I think this problem involves making use of symmetry in some way but I don't know how. I expanded the $(x+y)^4$ term but it din't help in the factorization. I am very bad at factorizing, so I didn't work too long on this problem. Any help would be appreciated, I want hints rather than a full solution. But even a full solution is okay.
 A: Note
\begin{align}
&x^4+y^4+(x+y)^4\\
=&(x^2+y^2)^2 -2x^2y^2 + (x^2+y^2+2xy)^2\\
=&2(x^2+y^2)^2 +4xy(x^2+y^2)+ 2x^2y^2\\
=&2(x^2+y^2+xy)^2
\end{align}
A: If $a+b+c=0$
$$a^4+b^4+c^4$$
$$=(a^2+b^2)^2-2a^2b^2+c^4$$
$$=((a+b)^2-2ab)^2-2a^2b^2+c^4$$
$$=((-c)^2-2ab)^2-2a^2b^2+c^4$$
$$=2(c^2-ab)^2$$
Here $a=-x,b=-y,c=?$
A: Let $s={x+y\over 2}$ so $x,s,y$ make an arithmetic progression with difference name it $d$. So $x=s-d$ and $y=s+d$. 
Now \begin{align}E&= (s-d)^4+(s+d)^4+16s^4\\
&= 2s^4+12s^2d^2+2d^4+16s^4\\ 
&= 18s^4+12s^2d^2+2d^4\\ 
&=2(9s^4+6s^2d^2+d^4)\\
&=2(3s^2+d^2)^2
\end{align}
A: $x^4+y^4+(x+y)^4=2x^4+6x^2y^2+4x^3y+2y^4+4y^3x=2(\color{green}{x^4}+3\color{green}{x^2y^2}+2x^3y+\color{green}{y^4}+2y^3x)$
You can see by expanding that $3$ terms are perfect squares, so you can attempt $(x^2+a.xy+y^2)^2$ and see what's left, with a coefficient for $xy$ since this is unsure how much $xy$ there is.
It is not guaranteed to works, but if the expression is indeed a square, your chances are high.
A: $$x^4+y^4+(x+y)^4$$
$$=2x^4+4x^3y+6x^2y^2+4xy^3+2y^4$$
$$=2(x^4+2x^3y+3x^2y^2+2xy^3+y^4)$$
$$=2(x^2+xy+y^2)^2$$
The last bit I guessed from $111^2=12321$.
