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.
 A: 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.
A: 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?
A: 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 $1\le k\le 5$.
A: Expand out the right hand side (binomial theorem will save you time):
$$x^6=x^6+6x^5+15x^4+20x^3+15x^2+6x+1$$ Subtract $x^6$ from both sides:
$$0=6x^5+15x^4+20x^3+15x^2+6x+1$$ Now just find the roots of this $5$th degree polynomial. If you are working over the complex numbers there will be exactly $5$ (see fundamental theorem of algebra), if you are only working in the real numbers $x=-\frac 12$ is the only solution.
