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

Solution 1:

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 $1\le k\le 5$.

Solution 2:

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.

Solution 3:

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?