Proving $\text{Li}_3\left(-\frac{1}{3}\right)-2 \text{Li}_3\left(\frac{1}{3}\right)= -\frac{\log^33}{6}+\frac{\pi^2}{6}\log 3-\frac{13\zeta(3)}{6}$? Ramanujan gave the following identities for the Dilogarithm function:
$$
\begin{align*}
\operatorname{Li}_2\left(\frac{1}{3}\right)-\frac{1}{6}\operatorname{Li}_2\left(\frac{1}{9}\right) &=\frac{{\pi}^2}{18}-\frac{\log^23}{6} \\
\operatorname{Li}_2\left(-\frac{1}{3}\right)-\frac{1}{3}\operatorname{Li}_2\left(\frac{1}{9}\right) &=-\frac{{\pi}^2}{18}+\frac{1}{6}\log^23 
\end{align*}
$$
Now, I was wondering if there are similar identities for the trilogarithm? I found numerically that 
$$\text{Li}_3\left(-\frac{1}{3}\right)-2 \text{Li}_3\left(\frac{1}{3}\right)\stackrel?= -\frac{\log^3 3}{6}+\frac{\pi^2}{6}\log 3-\frac{13\zeta(3)}{6} \tag{1}$$


*

*I was not able to find equation $(1)$ anywhere in literature. Is it a new result? 

*How can we prove $(1)$? I believe that it must be true since it agrees to a lot of decimal places.

 A: By "anywhere in literature" did you include L. Lewin, Polylogarithms and Assoicated Functions ?
In the (fortcoming) solution for Monthly problem 11654, you will see that very equation.  For the proof, our solver (Richard Stong) used these identities:
$$
\mathrm{Li}_3\left(\frac{1-z}{1+z}\right) - \mathrm{Li}_3\left(\frac{z-1}{z+1}\right) =
2\;\mathrm{Li}_3(1-z) + 2\;\mathrm{Li}_3\left(\frac{1}{z+1}\right) -
\frac{1}{2}\;\mathrm{Li}_3(1-z^2) - \frac{7}{4}\zeta(3)-
\frac{1}{3}\big(\log(1+z)\big)^3 + \frac{\pi^2}{6}\log(1+z)
$$
at $z=2$ and
$$
\mathrm{Li}_3(z) = \mathrm{Li}_3\left(\frac{1}{z}\right) -
 \frac{1}{6}\big(\log(-z)\big)^3 - \frac{\pi^2}{6}\log(-z)
$$
at $z=-3$.  Stong cites the Lewin book.
A: Combining trilogarithm identities 1 and 2, one obtains the formula
\begin{align}
\operatorname{Li}_3\left(\frac{1-z}{1+z}\right)-\operatorname{Li}_3\left(-\frac{1-z}{1+z}\right)=
2\operatorname{Li}_3\left(1-z\right)+2\operatorname{Li}_3\left(\frac{1}{1+z}\right)-
\frac12\operatorname{Li}_3\left(\frac{1}{1-z^2}\right)-\frac74\operatorname{Li}_3\left(1\right)\\
-\frac13\ln^3(z+1)+\frac{\pi^2}{6}\ln(z+1)+\frac{1}{12}\ln^3\left(z^2-1\right)+\frac{\pi^2}{12}\ln(z^2-1).
\end{align}
Now it suffices to set $z=2$ and use that $\operatorname{Li}_3\left(1\right)=\zeta(3)$, $\operatorname{Li}_3\left(-1\right)=-\frac34\zeta(3)$.
