TITLE: solve equation $y^{\alpha} + y^{1 + \alpha} = x $ 
QUESTION [6 upvotes]: how to solve this equation:
$y^{\alpha} + y^{1 + \alpha} = x $ where $\alpha \in (-1, 0)$
Is there trick to solve it?
EDIT. I want to find $y(x)$.

REPLY [4 votes]: Lagrange inversion formula, more precisely Lagrange-Bürmann formula is made for this.
The function $y$ solves $y=z\phi(y)$, with $z=x^{1/\alpha}$ and $\phi:t\mapsto(1+t)^{-1/\alpha}$. Since $\phi(0)\ne0$, one knows that a solution is
$$
y(x)=\sum_{n\geqslant1}a_nz^n=\sum_{n\geqslant1}a_nx^{n/\alpha},\qquad a_n=\frac1n[t^{n-1}]\phi(t)^n.
$$
In the present case, for every $\alpha$ in $(-1,0)$,
$$
a_n=\frac1n{-n/\alpha\choose n-1}=-\frac{\alpha}n{-n/\alpha+1\choose n},
$$
and one can check that the resulting series converges for every $x$ such that $|x|\gt R_\alpha$ with
$$
R_\alpha=\frac{(-\alpha)^\alpha}{(1+\alpha)^{1+\alpha}}.
$$
Sanity check: If $\alpha=-\frac12$, one gets $R_\alpha=2$, as was to be expected since an explicit formula for $y(x)$ in this case is
$$
y(x)=-1+\frac12x^2-\frac12x^2\sqrt{1-\frac4{x^2}}=\frac{1-\sqrt{1-\frac4{x^2}}}{1+\sqrt{1-\frac4{x^2}}}.
$$
Edit:
For every $\beta\gt0$ and $\gamma$, when $n\to\infty$,
$$
\frac{\Gamma(\beta(n+1)+\gamma)}{\Gamma(\beta n+\gamma)}\sim \beta^\beta n^\beta.
$$
Applying this to each Gamma function in the expression of 
$$
a_n=\dfrac{\Gamma(-n/\alpha+1)}{\Gamma(n+1)\Gamma(-n/\alpha-n+2)},
$$
one sees that the powers of $n$ cancel out and one is left with
$$
\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=\frac{(-1/\alpha)^{-1/\alpha}}{1^1\,(-1/\alpha-1)^{-1/\alpha-1}}=(-1/\alpha)\,(1+\alpha)^{(1+\alpha)/\alpha}=\varrho(\alpha).
$$
Hence the series $y(x)$ converges for every $x$ such that $\varrho(\alpha)|x|^{1/\alpha}\lt1$ and diverges for every $x$ such that $\varrho(\alpha)|x|^{1/\alpha}\gt1$. Since $\alpha\lt0$, the first condition reads $|x|\gt\varrho(\alpha)^{-\alpha}$ and the second condition reads $|x|\lt\varrho(\alpha)^{-\alpha}$. Since $\varrho(\alpha)^{-\alpha}=R_\alpha$, the proof is complete.