Inverse of $y= x^{3} +x+1$ Can you help me find the inverse function for  $y= x^3 +x+1$? This question proposed by one of my friends and I don't know the real source of the problem.
We can't use Cardano's method for solve this problem.

Thanks!
Why won't Cardano work?



I'll tackle this problem the algebraic way because it's the most straightforward, despite the intimidating algebra.

Replacing $x$ with $y$, we get$$y^3+y+1=x\quad\implies\quad y^3+y+1-x=0$$Now recall Cardano's formula for the cubic $x^3+qx+r=0$ who has the real root at$$x_1=\left\{-\frac r2+\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}+\left\{-\frac r2-\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}$$Substituting $q=1$ and $r=1-x$ gives$$\begin{align*}y=\sqrt[3]{-\frac {1-x}2+\sqrt{\frac {27x^2-54x+31}{108}}}-\sqrt[3]{\frac {1-x}2+\sqrt{\frac {27x^2-54x+31}{108}}}\end{align*}$$And that's your inverse! Here's a visual diagram in Desmos
Given

$$ y = x^3+x+1 $$

Interchange $x,y$ to see inverse function graph. 

$$ x= y^3+y+1 $$

Using Cardano it is quite possible to find the solution $f(x)$ in terms of $y$ analytically. There is one real root at $x=1$ coinciding at its inflection point.