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.

Hint We may factor this as:
$$(x-(x+1))(x+(x+1))(x^2-x(x+1)+(x+1)^2)(x^2+x(x+1)+(x+1)^2)=0$$
So then we have a linear and two quadratics which you should be able to solve.
