<s>[INST] Inverse of $y = Ax^3 + Bx + C$ I know there already is a question and a bunch of excellent answers in How to find the inverse of $y=x^3-5x^2+3x+c$.
But here I am having a more general (and thus complicated) form to invert. I am stuck in obtaining the inverse, $g^{-1}$, of a polynomial
\begin{align}
y = g\left(x\right) = -\frac{x^3}{8r} - r\left(1 + \frac{1}{\sqrt{2}}\right)x + 2\sqrt{2}r^2, {\rm{~~~~}} 0 \le x \le \sqrt{2}r
\end{align}
where $r > 0$ is a constant and a real number.
The function $g$ is certainly invertible as being strictly decreasing when drawn:

Question:
(1) I would like to ask you guys if there would be a way to find a closed-form expression for $x = g^{-1}\left(y\right)$.
(2) Also, any approximation of $y$ for making it easier to be inverted will be most welcome as well.
Thanks a lot guys. I love you all!
 [/INST] Consider that you need to solve for $x$ the cubic equation
$$ -\frac{x^3}{8r} - r\left(1 + \frac{1}{\sqrt{2}}\right)x + \left(2\sqrt{2}r^2-y\right)=0$$
Applying the method for cubic equations, you would find
$$\Delta=-\frac{27 \left(2 \sqrt{2} r^2-y\right)^2}{64 r^2}-\frac{1}{2}
   \left(1+\frac{1}{\sqrt{2}}\right)^3 r^2$$ and, since $\Delta <0$, then only one real root. On the other hand $$ p=8 \left(1+\frac{1}{\sqrt{2}}\right) r^2\qquad \text{and}\qquad q=-8 r \left(2 \sqrt{2} r^2-y\right)$$ 
Using the hyperbolic solution for one real root
$$x=4 \sqrt{\frac{2+\sqrt{2}}{3} }  \sinh \left(\frac{1}{3} \sinh
   ^{-1}\left(\frac{3\sqrt{3} \left(2 \sqrt{2} r^2- y\right)}{2
   \left(2+\sqrt{2}\right)^{3/2} r^2}\right)\right)\,r$$
Edit
It would have been simpler to let $x= X r \sqrt 2$ and $y= 2Y {r^2}\sqrt 2$ which make the equation to be dimensionless and rescaled
$$\sqrt{2} X^3+4 \left(1+\sqrt{2}\right) X-8 \sqrt{2} (1-Y)=0$$
the solution of which being
$$X=2 \sqrt{\frac{2\left(2+\sqrt{2}\right)}{3} } \sinh \left(\frac{1}{3} \sinh
   ^{-1}\left(\frac{3 \sqrt{6} }{\left(2+\sqrt{2}\right)^{3/2}}(1-Y)\right)\right)$$
</s>