<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] $$f(x)=x^8+(x+2)^8\qquad f'(x)=8x^7+8(x+2)^7\qquad f''(x)=56x^6+56(x+2)^6$$
$f''(x)$is clearly positive so $f$ is convex and above its tangents, in particular since $f'(-1)=0$ it is above the horizontal line $y=f(-1)+(x-1)f'(-1)=2$.
Since this value is effectively reached for $x=-1$ then it is the global minimum (no other because of convexity), and you can generalize to any $2n$ exponent instead of $8$.
</s>