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

$x^3 - \frac {x}{b} =-1$
Let $x = 2\frac {\sqrt {3b}}{3b}\cos\theta$
$\frac {8\sqrt {3b}}{9b^2} \cos^3\theta - \frac {2\sqrt{3b}}{3b^2}\cos\theta = 1\\
\frac {2\sqrt{3b}}{9b^2}(4\cos^3\theta - 3\cos\theta) = 1\\
\cos3\theta = \frac {3b\sqrt {3b}}{2}\\
\theta = \frac 13 \arccos\left(\frac {3b\sqrt {3b}}{2}\right), \frac 13 \arccos\left(\frac {3b\sqrt {3b}}{2}\right) + \frac {2\pi}{3}, \frac 13 \arccos\left(\frac {3b\sqrt {3b}}{2}\right) + \frac {4\pi}{3} \\
x =  2\frac {\sqrt {3b}}{3b}\cos\left(\frac 13 \arccos\left(\frac {3b\sqrt {3b}}{2}\right)\right), 2\frac {\sqrt {3b}}{3b}\cos\left(\frac 13 \arccos\left(\frac {3b\sqrt {3b}}{2}\right)+\frac {2\pi}{3}\right), 2\frac {\sqrt {3b}}{3b}\cos\left(\frac 13 \arccos\left(\frac {3b\sqrt {3b}}{2}\right)+\frac {4\pi}{3}\right)$
