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.

We can simplify the problem by substituting $x=y-\frac{1}{2}$.
$$x^6 = \left(x+1\right)^6$$
$$\left(y-\frac{1}{2}\right)^6 = \left(y+\frac{1}{2}\right)^6$$
If we expand both sides and then collect the terms, the even powers of $y$ drop out and only the odd powers remain.
$$6y^5+5y^3+\frac{3}{8}y=0$$
You can factor out the root at $y=0$ (i.e. corresponding to the solution $x=-\frac{1}{2}$) and then factor the remaining polynomial into two quadratics.
$$6y\left(y^2+\frac{3}{4}\right)\left(y^2+\frac{1}{12}\right)=0$$
Think you can take it from here?
