Factorise $x^4+y^4+(x+y)^4$

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}

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 $a=-x,b=-y,c=?$

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}