Complex solutions of $(x+1)^{10}+(x-1)^{10}=0$ Solve $$(x+1)^{10}+(x-1)^{10}=0$$

What method is available to solve this equation?
As pointed out by Gerry Myerson's comment , put $\frac{x+1}{x-1}=z$ for some complex $z,x\neq1$ Then, $x=\frac{z+1}{z-1}$. Now your equation reduces to $z^{10}=-1$, which is easily solvable by using Euler's/De-Moivre's Theorem as $z=e^{i\phi},\phi=\frac{(2k+1)\pi}{10},k\in\mathbb{Z^+\cup 0}:k&lt;10$ from which $x$ can be calculated
For $x\in \mathbb{R}$ $(x+1)^{10}+(x-1)^{10}=0$ don't have any solution. This is true by the argument Dr.Sonnhard Graubner.

So let's rewrite question, $$(z+1)^{10}+(z-1)^{10}=0\text{ for }z\in \mathbb{C}.$$

Then suppose if $(z-1)^{10}=0,$then $\implies z=1$ , but $(1+1)^{10}\not=0$ hence $(z-1)^{10}\not=0$.
\begin{gather}
\Rightarrow\frac {(z+1)^{10}}{(z-1)^{10}}+1=0 \\ \frac{z+1}{z-1}=\sqrt[10]{-1}
\\
\end{gather}
Hence $z=\frac{1+\sqrt[10]{(-1)}}{\sqrt[10]{(-1)} -1}$.