I would like to find a three-term asymptotic expansion of $g$, the inverse function of:
$ f(x)=x^3+x$
We have:
$ f(g(x))=x=g(x)^3+g(x)$ As $\ g(x) \rightarrow_{x\rightarrow \infty} \infty$
$ x=g(x)^3+g(x) \sim g(x)^3$
$ g(x) \sim x^{1/3}$
$ g(x)=x^{1/3}+o(x^{1/3})$
Moreover
$ g(x)=(x-g(x))^{1/3}=(x-x^{1/3}+o(x^{1/3}))^{1/3}$
$ g(x)=x^{1/3}(1-\frac{1}{x^{2/3}}+o(1/x^{2/3}))^{1/3}$
$ g(x)=x^{1/3}(1-\frac{1}{3x^{2/3}}+o(1/x^{2/3}))$
$ g(x)=x^{1/3}-\frac{1}{3x^{1/3}}+o(1/x^{2/3}) $
$ g(x)=(x-x^{1/3}+\frac{1}{3x^{1/3}}+o(\frac{1}{x^{1/3}}))^{1/3}$
$ g(x)=x^{1/3}(1-\frac{1}{x^{2/3}}+\frac{1}{3x^{4/3}}+o(\frac{1}{x^{4/3}}))^{1/3}$
$ g(x)=x^{1/3}(1-\frac{1}{3x^{2/3}}+\frac{1}{9x^{4/3}}-\frac{1}{9}(-\frac{1}{x^{2/3}}+\frac{1}{3x^{4/3}})^2+o(\frac{1}{x^{4/3}}))$
$ g(x)=x^{1/3}(1-\frac{1}{3x^{2/3}}+\frac{1}{9x^{4/3}}-\frac{1}{9x^{4/3}}+o(\frac{1}{x^{4/3}}))$
$ g(x)=x^{1/3}(1-\frac{1}{3x^{2/3}}+o(\frac{1}{x^{4/3}}))$
$ g(x)=x^{1/3}-\frac{1}{3x^{1/3}}+o(\frac{1}{x}) $
A solution is:
$ g(x)=(x-x^{1/3}+\frac{1}{3x^{1/3}}+o(\frac{1}{x}))^{1/3}$
$ g(x)=x^{1/3}(1-\frac{1}{x^{2/3}}+\frac{1}{3x^{4/3}}+o(1/x^2))^{1/3}$
$ g(x)=x^{1/3}(1-\frac{1}{3x^{2/3}}+\frac{1}{9x^{4/3}}-\frac{1}{9}(-\frac{1}{x^{2/3}}+\frac{1}{3x^{4/3}})^2+\frac{5}{81}(-\frac{1}{x^{2/3}}+\frac{1}{3x^{4/3}})^3)$
$ g(x)=x^{1/3}(1-\frac{1}{3x^{1/3}}+\frac{1}{9x^{4/3}}-\frac{1}{9x^{4/3}}+\frac{2}{27x^2}-\frac{5}{81x^2})$
$ g(x)=x^{1/3}-\frac{1}{3x^{1/3}}+\frac{1}{81x^{5/3}}+o(1/x^{5/3})$