Simplification of an expression containing $\operatorname{Li}_3(x)$ terms In my computations I ended up with this result:
$$\mathcal{K}=78\operatorname{Li}_3\left(\frac13\right)+15\operatorname{Li}_3\left(\frac23\right)-64\operatorname{Li}_3\left(\frac15\right)-102
\operatorname{Li}_3\left(\frac25\right)+126\operatorname{Li}_3\left(\frac35\right)\\+12\operatorname{Li}_3\left(\frac45\right)-89\operatorname{Li}_3\left(\frac16\right)-152\operatorname{Li}_3\left(\frac56\right)+63\operatorname{Li}_3\left(\frac38\right)+76\operatorname{Li}_3\left(\frac58\right).$$
I wonder if it's possible to simplify this expression somehow, e.g. to combine some trilogarithm terms to express them using logarithms, or at least to reduce the number of terms?
I tried to apply identities given at MathWorld and Wolfram Functions, but could not make the overall expression simpler. Mathematica could not simplify it either.

Too long for a comment, but a related thing, really far from the solution.
$$\operatorname{Li}_3\left(\frac{3}{8}\right)+\operatorname{Li}_3\left(\frac{5}{8}\right)-\operatorname{Li}_3\left(\frac{3}{5}\right) = \frac{1}{6} \ln^3\left(\frac{5}{8}\right)-\frac{1}{2}\ln\left(\frac{3}{8}\right)\ln^2\left(\frac{5}{8}\right) + \frac{\pi^2}{6}\ln\left(\frac{5}{8}\right) + \zeta(3) - \frac{1}{4} \operatorname{Li}_3\left(\frac{9}{25}\right).$$
I get it by using this identitiy for $z:=3/8$.
An other one for $z:=1/6$ with the same identity.
$$\operatorname{Li}_3\left(\frac{1}{6}\right)+\operatorname{Li}_3\left(\frac{5}{6}\right)-\operatorname{Li}_3\left(\frac{1}{5}\right) = \frac{1}{6} \ln^3\left(\frac{5}{6}\right)-\frac{1}{2}\ln\left(\frac{1}{6}\right)\ln^2\left(\frac{5}{6}\right) + \frac{\pi^2}{6}\ln\left(\frac{5}{6}\right) + \zeta(3) - \frac{1}{4} \operatorname{Li}_3\left(\frac{1}{25}\right).$$
