How to find the inverse of the function $(1/4)x^3+x-1$

Question

I am hoping someone can help me find the inverse of the function $(1/4)x^3+x-1$. I first switched $x$ and $y$ in the equation to get $x=(1/4)y^3+y-1$. However, after this stage I was completely stuck trying to solve for $y$, as each time there was always two $y$’s on one side. Also the question is asking me to find $g^{-1}(x)$ when $x=3$, which I know after I find the inverse I have to substitute $3$ into the equation to get the value. Thanks for any insight!

Answer 1

Apply Cardano's formula to $x^3 + 4x - (4 + 4y) = 0$.

Answer 2

If you just want to find $g^{-1}(3)$ note that $$\frac{1}{4}x^3+x-1=3$$ $$\implies x^3+4x-16=0$$ and a solution to this is $x=2$