Inverse function of a polynomial $x(x-1)(x+1) = 0 $ How to get the interval to get the inverse function of the polynomial $$x(x-1)(x+1) = f(x) $$, to show that the function is surjective.
The interval where the function is surjective is $x > 1$ and $x < -1$, how to get inverse on this interval ?
 A: Note that
$$x= y(y-1)(y+1)=y^3-y$$
is cubic in $y$. Use Cardano’s formula to obtain the inverse function below
$$y= \sqrt[3]{\frac x2+\sqrt{\frac{x^2}4-\frac1{27}}} +\sqrt[3]{\frac x2-\sqrt{\frac{x^2}4-\frac1{27}}}
$$
with $x\ne 0$.
A: This is the overkill approach.

Theorem: (Lagrange Inversion Theorem)
Suppose $z$ is defined as a function of $w$ by an equation of the form $$z=f(w),$$
where $f$ is analytic at a point $a$, and $f'(a)\ne 0$. Then we may solve for $w$ as
$$w=a+\sum_{n\ge1}\frac{q_n}{n!}(z-f(a))^n,\tag1$$
where $$q_n=\lim_{y\to a}\left[\left(\frac{\partial}{\partial y}\right)^{n-1}\left(\frac{y-a}{f(y)-f(a)}\right)^n\right].\tag 2$$

We may apply this theorem to the function $$z=f(w)=w(w+1)(w-1)=w^3-w.$$
Clearly $f$ is analytic. We can choose $a=0$ because $f'(0)=-1$. Thus
$$w=f^{-1}(z)=\sum_{n\ge1}\frac{q_n}{n!}z^n,$$
with $$q_n=\lim_{y\to 0}\left[\left(\frac{\partial}{\partial y}\right)^{n-1}\frac{1}{(y^2-1)^n}\right].$$
Then, recall that for  we have
$$\frac{1}{(1-q)^\alpha}=\sum_{k\ge0}\frac{(\alpha)_k}{k!}q^k,$$
where $(\alpha)_k=\Gamma(\alpha+k)/\Gamma(\alpha)$ and $\alpha\in\Bbb R$. Then,
$$\frac{1}{(y^2-1)^n}=\sum_{k\ge0}(-1)^n\frac{(n)_{k/2}}{(k/2)!}[2|k]y^k,$$
where
$$[2|k]=
\begin{cases}
1 & \text{if } 2\text{ divides }k\\
0 & \text{otherwise}
\end{cases}
= \frac{1+(-1)^k}{2}.
$$
Letting $p_n(y)=1/(y^2-1)^n$, we have the Taylor series of $p_n$ is
$$p_n(y)=\sum_{k\ge0}\frac{p_n^{(k)}(0)}{k!}y^k,$$
thus $$p_n^{(k)}(0)=(-1)^n\frac{(n)_{k/2}k!}{(k/2)!}[2|k].$$
And since $q_n=p_n^{(n-1)}(0)$, we have
$$q_n=(-1)^n\frac{(n)_{(n-1)/2}(n-1)!}{((n-1)/2)!}[2|n-1],$$
so that $$f^{-1}(z)=\sum_{n\ge1}(-1)^n\frac{(n)_{(n-1)/2}(n-1)!}{n!((n-1)/2)!}[2|n-1]z^n=\sum_{n\ge1}\frac{(-1)^n}{n}\frac{(n)_{(n-1)/2}}{((n-1)/2)!}[2|n-1]z^n.$$
This result is mess, so lets clean it up. Preforming the index shift $j=n-1$, we have
$$\begin{align}
f^{-1}(z)&=\sum_{j\ge0}\frac{(-1)^{j+1}}{j+1}\frac{(j+1)_{j/2}}{(j/2)!}[2|j]z^{j+1}\\
&=\sum_{j\ge0}\frac{(-1)^{2j+1}}{2j+1}\frac{(2j+1)_{j}}{(j)!}[2|2j]z^{2j+1}+\sum_{j\ge0}\frac{(-1)^{2j+2}}{2j+2}\frac{(2j+2)_{(2j+1)/2}}{((2j+1)/2)!}[2|2j+1]z^{2j+2}\\
&=-\sum_{j\ge0}\frac{(2j+1)_{j}}{(j)!}\frac{z^{2j+1}}{2j+1},\tag{3}
\end{align}$$
where we get $(3)$ from the fact that $[2|2j]=1$ and $[2|2j+1]=0$. Then using the definition of $(x)_n$, we have
$$f^{-1}(z)=-\sum_{n\ge0}\frac{\Gamma(3n+1)z^{2n+1}}{\Gamma(2n+2)\Gamma(n+1)}.$$
We are fortunate enough to express this as a hypergeometric series. To do so, write
$$t_n=\frac{\Gamma(3n+1)z^{2n+1}}{\Gamma(2n+2)\Gamma(n+1)}.$$
Then
$$\frac{t_{n+1}}{t_n}=\frac{(n+1/3)(n+2/3)}{n+3/2}\frac{27z^2/4}{n+1}, \qquad t_0=z.$$
Thus
$$f^{-1}(z)=-z\,_2F_1(\tfrac{1}{3},\tfrac{2}{3};\tfrac{3}{2};\tfrac{27}{4}z^2)\qquad |z|<\frac{2}{3\sqrt3}.$$

Edit: In case you're interested, here is some numerical verification of my answer.
A: Consider $$x(x-1)(x+1) =x^3-x= y$$ and then th cubic equation
$$x^3-x-y=0$$ for which the discriminant $\Delta=4-27y^2$ must be negative in order to have only one real root. This defines
$$-\frac{2}{3 \sqrt{3}} <y <\frac{2}{3 \sqrt{3}}$$
Using the hyperboic method for only one real root, we then have
$$x=\frac{2}{\sqrt{3} }\frac{ |y|}y   \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{3 \sqrt{3}
   }{2}|y|\right)\right)$$
