Expanding the inverse of x + x³

Setup. $g$ is differentiable with $g'(y)=\frac1{1+3g(y)^2}$, and it is infinitely differentiable, so it has expansions to any order. Also $g(y)=y+o(y)$ because $g'(0)=1$.

Step 1. From $g(y)=y-g(y)^3$ and $g(y)=y+o(y)$:
$$g(y)=y-y^3+o(y^3).$$

Step 2. Cube it: $g^3=\left(y-y^3+o(y^3)\right)^3=y^3-3y^5+o(y^5)$. Hence
$$g(y)=y-\left(y^3-3y^5\right)+o(y^5)=y-y^3+3y^5+o(y^5).$$

Check. With $y=0.1$: $y-y^3+3y^5=0.099\,03$ and $f(0.099\,03)=0.099\,03+0.000\,971\approx0.100\,00$ ✓.

Remark. Each substitution gains two orders of accuracy, because $g^3$ is $O(y^3)$. This bootstrap method works for any equation of the form $x=y+\varepsilon(x)$.