While reading about polylogarithms, I came across the nice polylogarithm ladder,
$$6\operatorname{Li}_2(x^{-1})-3\operatorname{Li}_2(x^{-2})-4\operatorname{Li}_2(x^{-3})+\operatorname{Li}_2(x^{-6}) = \frac{7\pi^2}{30}\tag{1}$$
where $x = \phi = \frac{1+\sqrt{5}}{2}$, the golden ratio, or the root $1<x<2$ of,
$$x^n(2-x) = 1\tag{2}$$
for the case $n=2$. I wondered if there was anything for $n=3$ (the tribonacci constant $T$, or the real root of $x^3-x^2-x-1 = 0$). After a little expt with Mathematica's integer relations command, I found,
$$4\operatorname{Li}_2(x^{-1})-4\operatorname{Li}_2(x^{-3})-3\operatorname{Li}_2(x^{-4})+\operatorname{Li}_2(x^{-8}) = \frac{\pi^2}{6}\tag{3}$$
However, for general $n$ (including the tetranacci constant and higher), it seems they obey,
$$4\operatorname{Li}_2(x^{-1})-\operatorname{Li}_2(x^{-2})+\operatorname{Li}_2(x^{-n+1})-2\operatorname{Li}_2(x^{-n}) = 2\operatorname{Li}_2(-x)+\frac{\pi^2}{2}\tag{4}$$
where $x$ is the root $1<x<2$ of $(2)$.
Q: Anybody knows how to prove if $(4)$ is indeed true for all integer $n\geq2$?