Zeros of $a(1-x)x^b-(1-a)x(1-x)^b = 0$ What are the zeros of this equation?

$$a(1-x)x^b-(1-a)x(1-x)^b = 0$$

I was told that there were 3 zeros, however I've only been able to find 2 of them. Namely, $x=0$ and $x=1$. I can't think of any way to solve it algebraically and I'm not seeing any other obvious solutions. I also tried using Taylor expansion on the $(1-x)^b$ term, but it didn't prove to be too helpful. I graphed the equation with an arbitrary choice of a and b, but only noticed 2 zeros. 

The equation has at either $3$ or $4$ roots, assuming that $0 < a < 1$.
In particular, the third root they were probably looking for is
$$
x = \frac{1}{1 + \sqrt[(b-1)]{r}},
$$
where $r = \frac{a}{1-a}$.
This root is between $0$ and $1$.
If $b$ is odd and $a \ne \frac12$, there is a second root,
$$
x = \frac{1}{1 - \sqrt[(b-1)]{r}}.
$$

Let us write
$$
f(x) = x(1-x)\left( a x^{b-1} - (1-a) (1-x)^{b-1} \right).
$$
So in addition to the roots $x = 0$ and $x = 1$, we will have a root if
$$
a x^{b-1} = (1-a) (1-x)^{b-1}
$$
i.e.,
$$
\left(\frac{1}{x} - 1\right)^{b-1} = \frac{a}{1 - a}.
$$
Now, $\frac{a}{1-a}$ is some positive number $r$.
So it has a $(b-1)$th root, $\sqrt[(b-1)]{r} > 0$.
So $\frac{1}{x} = 1 + \sqrt[(b-1)]{r} > 1$,
and
$$
\boxed{x = \frac{1}{1 + \sqrt[(b-1)]{r}}}.
$$
If $b$ is odd, $b-1$ is even, so there is also a negative $(b-1)$th root of $r$, and thus we also have the solution
$$
x = \frac{1}{1 - \sqrt[(b-1)]{r}}.
$$
