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
$$x = \sum_{k=0}^\infty \binom{3k+2}{k} \frac{b^{3k+2} }{(3)k+2} $$

Too long for a comment.
Looking for truncated series expansions, you can directly solve for the roots solving the cubic using the trigonometric method.
The solutions are given by
$$x_k=\frac{2 \cos \left(\frac{2 \pi  }{3}k-\frac{1}{3} \cos
   ^{-1}\left(-\frac{3\sqrt{3}}{2}  b^{3/2}\right)\right)}{\sqrt{b}\sqrt{3}}\qquad \text{with} \qquad k=0,1,2$$
We have
$$x_1=b+b^4+3 b^7+12 b^{10}+55 b^{13}+273 b^{16}+1428 b^{19}+7752
   b^{22}+O\left(b^{25}\right)$$ and the coefficients corrrespond to sequence $A001764$ in $OEIS$ that is to say
$$x_1=\sum_{n=0}^\infty \frac{ \binom{3 n}{n}}{2 n+1}b^{3 n+1}$$ which is the one you obtained using  Lagrange inversion theorem.
Similarly,
$$x_0=\frac{1}{\sqrt{b}}-\frac{b}{2}-\frac{3 b^{5/2}}{8}-\frac{b^4}{2}-\frac{105
   b^{11/2}}{128}-\frac{3 b^7}{2}-\frac{3003 b^{17/2}}{1024}-6
   b^{10}+O\left(b^{23/2}\right)$$
$$x_2=-\frac{1}{\sqrt{b}}-\frac{b}{2}+\frac{3 b^{5/2}}{8}-\frac{b^4}{2}+\frac{105
   b^{11/2}}{128}-\frac{3 b^7}{2}+\frac{3003 b^{17/2}}{1024}-6
   b^{10}+O\left(b^{23/2}\right)$$ already given by @Somos.
You also have $x_0+x_1+x_2=0$ and $x_0x_1x_2=-1$ also as mentioned by @Somos.
