Inverse a cubic function

Question

I've got this cubic function that I can't figure out how to calculate its inverse.

$$f(x)=x^3+3x^2+3x, x\in \mathbb{R}$$

I've tried using online calculator to see if that would help, but none of them seemed to solve this.

Answer 1

Set $y=x^3+3x^2+3x$, and notice that $(x+1)^3=x^3+3x^2+3x+1 \implies y=(x+1)^3-1.$

Now we can just rearrange a bit (with a cube root thrown in there) to note $x=\sqrt[3]{y+1}-1.$

Thus, if $f(x)=x^3+3x^2+3x,$ then $f^{-1}(x)=\sqrt[3]{x+1}-1.$