Find all $x$ such that $x^6 = (x+1)^6$. 
Find all $x$ such that $$x^6=(x+1)^6.$$

So far, I have found the real solution $x= -\frac{1}{2}$, and the complex solution $x = -\sqrt[3]{-1}$.
Are there more, and if so what/how would be the most efficient find all the solutions to this problem? I am struggling to find the rest of the solutions.

Since $0$ is not a solution, this is equivalent to solve
$$
\frac{x+1}{x}=\zeta
$$
where $\zeta$ is any sixth root of $1$.
Then $x+1=\zeta x$, so
$$
x=\frac{1}{\zeta-1}
$$
Of course $\zeta=1$ should not be considered. There are other five sixth root of $1$, namely
$$
e^{2k\pi i/3}
$$
for $1\le k\le 5$.
