Inverse of $x^3 + x +1$

Question

How to calculate the inverse of function

$y=x^3 + x + 1$

Tried this: https://www.symbolab.com/solver/function-inverse-calculator/inverse%20f%5Cleft(x%5Cright)%3Dx%5E%7B3%7D%2Bx%2B1

But this doesn't show the steps of solution.

Thanks!

Answer 1

$x^3 + x = y-1$ let $x = z - \frac {1}{3z}$

$(z - \frac {1}{3z})^3 + z - \frac {1}{3z} = y-1\\ z^3 - \frac {3z^2}{3z} + \frac {3z}{9z^2} - \frac {1}{27z^3} + z - \frac {1}{3z} = y-1\\ z^3- \frac {1}{27z^3} = y-1\\ z^6 - (y-1)z^3 - \frac {1}{27} = 0 \\ z^3 = \frac {y-1 \pm \sqrt {(y-1)^2 +\frac {4}{27}}}{2}$

If we say $z^3 = \frac {y-1 + \sqrt {(y-1)^2 +\frac {4}{27}}}{2}$ then $\frac {y-1 - \sqrt {(y-1)^2 +\frac {4}{27}}}{2} = -\frac {1}{27z^3}$

$x = \sqrt[3]{\frac {y-1 + \sqrt {(y-1)^2 +\frac {4}{27}}}{2}} + \sqrt[3]{\frac {y-1 - \sqrt {(y-1)^2 +\frac {4}{27}}}{2}} $