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.

Surprisingly, $\mathcal K$ can be expressed in elementary terms. Let, $$a = \ln 2\\ b=\ln 3\\ c=\ln 5$$
Then,
$$\mathcal{K}=\frac23(878 a^3 - 37 b^3 - 7 c^3) - 2 a^2 (202 b + 133 c) + 4 b^2 (-32 a + 19 c) + 3 c^2(13 a - 21 b) + 278 a b c - \frac23 \pi^2 (22 a - 50 b + 25 c) \approx -7.809651$$
