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.


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 $\operatorname{Li}_3\left(1\right)=\zeta(3)$, $\operatorname{Li}_3\left(-1\right)=-\frac34\zeta(3)$.
