Factor Equations Please check my answer in factoring this equations:
Question 1. Factor $(x+1)^4+(x+3)^4-272$.
Solution: $$\begin{eqnarray}&=&(x+1)^4+(x+3)^4-272\\&=&(x+1)^4+(x+3)^4-272+16-16\\
&=&(x+1)^4+(x+3)^4-256-16\\
&=&\left[(x+1)^4-16\right]+\left[(x+3)^4-256\right]\\
&=&\left[(x+1)^2+4\right]\left[(x+1)^2-4\right]+\left[(x+3)^2+16\right]\left[(x+3)^2-16\right]\\
&=&\left[(x+1)^2+4\right]\left[(x+1)^2-4\right]+\left[(x+3)^2+16\right]\left[(x+3)-4\right]\left[(x+3)+4\right]\end{eqnarray}.$$
Question 2. Factor $x^4+(x+y)^4+y^4$
Solution: $$\begin{eqnarray}&=&(x^4+y^4)+(x+y)^4\\
&=&(x^4+y^4)+(x+y)^4+2x^2y^2-2x^2y^2\\
&=&(x^4+2x^2y^2+y^4)+(x+y)^4-2x^2y^2\\
&=&(x^2+y^2)^2+(x+y)^4-2x^2y^2
\end{eqnarray}$$
I am stuck in question number 2, I dont know what is next after that line.
 A: \begin{equation}
\begin{split}
\ & x^4+y^4+(x+y)^4\\
\ =& (x^2+y^2)^2-2x^2y^2+(x^2+y^2+2xy)^2\\
\ =& (x^2+y^2)^2-2x^2y^2+(x^2+y^2)^2+4xy(x^2+y^2)+4x^2y^2\\
\ =& 2((x^2+y^2)^2+x^2y^2+2xy(x^2+y^2))\\
\ =& 2(x^2+y^2+xy)^2
\end{split}
\end{equation}
A: \begin{align*}
(x+y)^4+x^4+y^4&=2(x^4+2x^3y+3x^2y^2+2xy^3+y^4)\\
&=2(x^4+2x^3y+2x^2y^2+x^2 y^2+2 x y^3+y^4)\\
&=2(x^4+2(xy+y^2)x^2+(xy+y^2)^2)\\
&=2(x^2+xy+y^2)^2
\end{align*}
A: For the first, I will put $y=\frac{x+1+x+3}2=x+2$
so that $x +1=y-1, x+3=y+1$ and the odd powers of $y$ vanish in $(y-1)^4+(y+1)^4$
$$\implies (x+1)^4+(x+3)^4-272=(y-1)^4+(y+1)^4-272$$
$$=2\{y^4+6y^2+1\}-272=2(y^4+6y^2-135)$$
$$=2\{y^4+(15-9)y^2-135\}=2(y^2+15)(y^2-9) =2(y^2-15)(y+3)(y-3)$$
$$=2\{(x+2)^2-15\}(x+5)(x-1)$$
$$\text{Now,  }(x+2)^2-15=x^2+4x+4-15=x^2+4x-11\text{ which is not reducible}$$ 
A: Let $w = x + 2$
\begin{align}
   (x+1)^4+(x+3)^4-272
   &= (w-1)^4+(w+1)^4-272\\
   &= 2w^4 + 12w^2 - 270\\
   &= 2(w^4 + 6w - 135)\\
   &= 2(w^4 +6w^2 + 9 - 144)\\
   &= 2(w^2 + 3)^2 - 12^2)\\
   &= 2(w^2 + 15)(w^2 - 9)\\
   &= 2(w^2 + 15)(w - 3)(w + 3)\\
   &=2(x^2 + 4x + 19)(x - 1)(x + 5)\\
\end{align}
\begin{align}
   (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^4 + x^3y + x^2y^2) +
                 (x^3y + x^2y^2 + xy^3) +
                         (x^2y^2 + xy^3 + y^4))\\
      &=   2(x^2(x^2 + xy + y^2) +
          xy(x^2 + xy + y^2) +
         y^2(x^2 + xy + y^2))\\
      &= 2(x^2 + xy + y^2)^2
\end{align}
