Point of inflexion of $y = \frac{9}{x + 1} - \frac{1}{x - 1}$ "31. Given that
$$y = \frac{9}{x + 1} - \frac{1}{x - 1},$$
(i) obtain expressions for $\dfrac{\text{d}y}{\text{d}x}$ and $\dfrac{\text{d}^{2}y}{\text{d}x^{2}}$,
(ii) find the stationary values of $y$,
(iii) show that there is only one point of inflexion and that this occurs for a value of $x$ lying between 2 and 3..."
Now,
$\dfrac{dy}{dx} = \dfrac{1}{(x - 1)^{2}} - \dfrac{9}{(x + 1)^{2}}$
$\dfrac{d^{2}y}{dx^{2}} = \dfrac{18}{(x + 1)^{3}} - \dfrac{2}{(x - 1)^{3}}$
And the stationary values of $y$ are 2 and 8, which occure where $x = 2$ and $x = \frac{1}{2}$, respectively.
It's on part (iii) that I'm stuck. I decided to try letting $\dfrac{d^{2}y}{dx^{2}} = 0$ and solving for $x$. What I get is a horrible cubic:
$4x^{3} - 15x^{2} + 12x - 5 = 0$
According to Wolfram|Alpha this does, indeed, have only one real solution and that solution is between 2 and 3. Which is excellent. But I have no idea how to solve this cubic myself, or show either of those things by other means.
Should I perhaps try a different method, or is there a trick to solve this cubic which I'm missing?

$\frac{1}{\frac{1}{9}(x+1)^3}=\frac{1}{(x-1)^3} \\ \frac{1}{(\frac{1}{9^\frac{1}{3}}(x+1))^3}=\frac{1}{(x-1)^3} \\   \frac{1}{9^\frac{1}{3}}(x+1)=x-1 \\ \text{ LoL, fixed it myself! } \\ \text{ continuing my work } \\ \frac{1}{9^{\frac{1}{3}}}x-x=-1-\frac{1}{9^{\frac{1}{3}}} \\ x(\frac{1}{9^\frac{1}{3}}-1)=-1-\frac{1}{9^\frac{1}{3}} \\ x=\frac{-1-\frac{1}{9^\frac{1}{3}}}{\frac{1}{9^\frac{1}{3}}-1}=\frac{-9^\frac{1}{3}-1}{1-9^\frac{1}{3}} $
