Prove that $\int_0^1 \frac{{\rm{Li}}_2(x)\ln(1-x)\ln^2(x)}{x} \,dx=-\frac{\zeta(6)}{3}$ I have spent my holiday on Sunday to crack several integral & series problems and I am having trouble to prove the following integral

\begin{equation}
\int_0^1 \frac{{\rm{Li}}_2(x)\ln(1-x)\ln^2(x)}{x} \,dx=-\frac{\zeta(6)}{3}
\end{equation}

Using integration by parts, $u=\ln^2(x)$ and $dv=\displaystyle\frac{{\rm{Li}}_2(x)\ln(1-x)}{x} \,dx$, I manage to obtain that the integral is equivalent to
\begin{equation}
\int_0^1 \frac{{\rm{Li}}_2^2(x)\ln(x)}{x} \,dx
\end{equation}
where ${\rm{Li}}_2^2(x)={\rm{Li}}_2(x)^2$, square of dilogarithm of $x$.
Could anyone here please help me to prove the above integral preferably with elementary ways (high school methods/ not residue method)? Any help would be greatly appreciated. Thank you.
 A: Here's another approach similar to M.N.C.E.'s post,
We have, 
$\displaystyle \begin{align} \int_0^1 \frac{\operatorname{Li_2}(x)\log (1-x) \log^2 x}{x}\,dx &= -\sum\limits_{n=1}^{\infty}\frac{1}{n}\int_0^1x^{n-1}\operatorname{Li_2}(x)\log^2 x\,dx \\ &=  -\sum\limits_{n=1}^{\infty}\frac{1}{n}\sum\limits_{m=1}^{\infty} \frac{1}{m^2} \int_0^1 x^{m+n-1}\log^2 x\,dx \\ &= -2\sum\limits_{n,m=1}^{\infty} \frac{1}{nm^2(m+n)^3} \\ &= -\sum\limits_{n,m=1}^{\infty} \frac{1}{nm^2(m+n)^3}-\sum\limits_{n,m=1}^{\infty} \frac{1}{mn^2(m+n)^3} \\ &= -\sum\limits_{n,m=1}^{\infty} \frac{1}{n^2m^2(m+n)^2} = -\frac{1}{3}\zeta(6) \end{align}$ 
The last double summation is proved here.
A: Another approach:
By Cauchy product we have 
$$\operatorname{Li}_2(x)\ln(1-x)=-\sum_{n=1}^\infty\left(\frac{2H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac3{n^3}\right)x^n$$
Multiply both sides by $\frac{\ln^2x}{x}$ then integrate from $x=0$ to $1$ and use the fact that $\int_0^1 x^{n-1}\ln^2xdx=\frac{2}{n^3}$ 
we get
$$\int_0^1\frac{\operatorname{Li}_2(x)\ln(1-x)\ln^2x}{x}dx=-4\sum_{n=1}^\infty\frac{H_n}{n^5}-2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}+6\zeta(6)$$
By Euler identity we have $\sum_{n=1}^\infty\frac{H_n}{n^5}=\frac74\zeta(6)-\frac12\zeta^2(3)$ and in my previous solution above we proved $\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}=\zeta^2(3)-\frac13\zeta(6)$. By collecting these results, the answer follows.
