<s>[INST] How to graph and solve this equation? I'am trying to solve this equation.
\begin{equation}
x^{8}+(x+2)^{8}=2
\end{equation}
What I tried:
\begin{equation}g(x)=x^{8}+(x+2)^{8}\end{equation}
\begin{equation}
\begin{array}{l}
\text {  }\\
y=x+1
\end{array}
\end{equation}
\begin{equation}
(y+1)^{8}+(y-1)^{8}=2
\end{equation}
\begin{equation}
(y+1)^{8}=y^{8}+a_{7} y^{7}+a_{6} y^{6}+\cdots+a_{1} y+1
\end{equation}
\begin{equation}
(y-1)^{8}=y^{8}-a_{7} y^{7}+a_{6} y^{6}+\cdots-a_{1} y+1
\end{equation}
\begin{equation}
\begin{array}{l}
2\left(y^{8}+a_{6} y^{6}+a_{4} y^{4}+a_{2} y^{2}+1\right)=2 \Leftrightarrow \\
y^{8}+a_{6} y^{6}+a_{4} y^{4}+a_{2} y^{2}=0 \Leftrightarrow y^{2}=0 \Leftrightarrow y=0
\end{array}
\end{equation}
y=0, x= -1
Maybe there is another way of solving this...
 [/INST] Let's try
$$ 1 + 1 = 2.$$
$$ \left\{ 
\begin{aligned}
x^8=1\\
(x+2)^8=1 
\end{aligned}
\right.
$$
A solution is $x=-1.$

Actually, $x=-1$ is a double root:
Let $q(x) = x^8 + (x+2)^8 -2$. So we have $$q(x)=0.$$  We can factor q(x):
$$q(x) = 2 (x+1)^2 (127 + 258 x + 253 x^2 + 132 x^3 + 43 x^4 + 6 x^5 + x^6).$$
The equation $127 + 258 x + 253 x^2 + 132 x^3 + 43 x^4 + 6 x^5 + x^6=0$
has no real solutions.
</s>