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...

OK then, let's use calculus.
Consider the function $f(x) = x^8 + (2 + x)^8$. Note that $f(x) \to \infty$ as $x \to \pm \infty$, so the function must achieve a global minimum at a stationary point, i.e. where $f'(x) = 0$.
We have,
$$f'(x) = 8x^7 + 8(x + 2)^7.$$
This is $0$ if and only if
$$8x^7 + 8(x + 2)^7 = 0 \iff (x + 2)^7 = -x^7 \iff \left(\frac{x + 2}{x}\right)^7 = -1.$$
Note, for the above to work, we must observe that $x = 0$ is definitely not a solution. As we are considering only the reals, we can take the unique seventh root of both sides to obtain
$$\frac{x + 2}{x} = -1 \iff x + 2 = -x \iff x = -1.$$
That is, there exists one and only one stationary point for the whole function: at $x = -1$. Therefore, the global minimum of $f$ must be achieved only at this point. That is,
$$2 = f(-1) \le f(x)$$
for all $x \in \Bbb{R}$, with equality if and only if $x = -1$. This implies $f(x) = 2$ if and only if $x = -1$.
