<s>[INST] Find the second real root for cubic $x^3+1-x/b=0$ A cubic of the form $$x^3+1-x/b=0$$   has has three real roots
Using the Lagrange inversion theorem one of the roots is given by
$$x = \sum_{k=0}^\infty \binom{3k}{k} \frac{b^{3k+1} }{(2)k+1} $$
How do you find the second one? I cannot find any info online. I am looking for a simialr series solution. I suspect is is of the form
$$x = \sum_{k=0}^\infty \binom{3k+2}{k} \frac{b^{3k+2} }{(3)k+2} $$
 [/INST] The related OEIS sequences A206300 and
A224884 help to answer your question. Define
$$ a_n := \frac{2^{2n-1}}{n!}\frac{\Gamma(3n/2-1/2)}{\Gamma(n/2+1/2)}. \tag{1} $$
Now define
$$ y_2 := \sum_{n=0}^\infty -a_n z^{3n-1},\quad
y_3 := \sum_{n=0}^\infty\, (-1)^na_n z^{3n-1},\quad
y_1 := -y_2-y_3. \tag{2} $$
These are the three roots of the cubic $$ y^3 - y/z^2 + 4 = 0. \tag{3}$$
The power series expansions are
$$ y_1 = 4z^2 + 64z^8 + 3072z^{14} + 196608z^{20} +\dots, \tag{4} $$
$$ y_2 = z^{-1} - 2z^2 - 6z^5 - 32z^8 - 210z^{11} - \dots, \tag{5}$$
$$ y_3 = -z^{-1} - 2z^2 + 6z^5 - 32z^8 +210z^{11} - \dots. \tag{6}$$
The roots of the cubic
$$ x^3 + 1 - x/b = 0 \tag{7} $$ are the roots of the previous cubic
divided by $\,2^{2/3}\,$ and where $\,z = b^{1/2}/2^{2/3}.$
The power series expansions are
$$ x_1 = b + b^4 + 3b^7 + 12b^{10} + 55b^{13} + \dots, \tag{8} $$
$$x_2 = b^{-1/2}-\frac12 b - \frac38 b^{5/2} - \frac12 b^4+\dots, \tag{9}$$
$$x_3 =-b^{-1/2}-\frac12 b+\frac38 b^{5/2}-\frac12 b^4+\dots. \tag{10}$$
Note that the coefficient of $\,x^{3n+1}\,$ in $\,x_1\,$ is
OEIS sequence A001764.
Note that $\,x_1 = b(1+x_1^3).\,$ Use it iteratively to generate
power series truncations of $\,x_1.$
Note that
$$ (x-x_1)(x-x_2)(x-x_3) = x^3-x/b+1, \tag{11}$$
which implies
$$ x_1+x_2+x_3 = 0,\quad x_1 x_2 x_3 = -1. \tag{12}$$
Thus, given a value for $\,x_1\,$ the other two roots $\,x_2,x_3\,$
satisfy a quadratic.
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