<s>[INST] 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.
 [/INST] 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 $(y-1)^4+(y+1)^4$
$$\implies (x+1)^4+(x+3)^4-272=(y-1)^4+(y+1)^4-272$$
$$=2\{y^4+6y^2+1\}-272=2(y^4+6y^2-135)$$
$$=2\{y^4+(15-9)y^2-135\}=2(y^2+15)(y^2-9) =2(y^2-15)(y+3)(y-3)$$
$$=2\{(x+2)^2-15\}(x+5)(x-1)$$
$$\text{Now,  }(x+2)^2-15=x^2+4x+4-15=x^2+4x-11\text{ which is not reducible}$$ 
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