Trying to prove that $\zeta(3)=\frac{5}{2} \sum_{k=1}^{\infty}{\frac{ (-1)^{k-1}} {\binom {2k}{k}k^{3}}}$

Creative telescoping is a very nice way for proving such identity, which is also discussed in the first section of my notes. On the other hand, one might also start with the RHS:

$$ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n^3\binom{2n}{n}} = \sum_{n\geq 1}\frac{(-1)^{n+1}B(n,n)}{2n^2}=\int_{0}^{1}\sum_{n\geq 1}\frac{(-1)^{n+1}}{2n^2}x^{n-1}(1-x)^{n-1}\,dx $$ turning the series in the LHS into $$ -\frac{1}{2}\int_{0}^{1}\frac{\text{Li}_2(-x(1-x))}{x(1-x)}\,dx\stackrel{x\mapsto\frac{1+z}{2}}{=}-\int_{-1}^{1}\frac{\text{Li}_2\left(-\frac{1-z^2}{4}\right)}{1-z^2}\,dz=-\int_{0}^{1}\frac{-\text{Li}_2\left(-\frac{x}{4}\right)}{x\sqrt{1-x}}\,dx.$$ By integration by parts, the RHS depends on $$ \int_{0}^{1}\log\left(1+\frac{1-x^2}{4}\right)\log\left(\frac{1-x}{1+x}\right)\,\frac{dx}{x} $$ which by enforcing the substitution $x=\frac{1-z}{1+z}$ is mapped into $$ \int_{0}^{1}\frac{\log(z)\log\left(\frac{1+3z+z^2}{1+z}\right)}{1-z^2}\,dz. $$ The last integral can be computed from the functional relations for $\text{Li}_2$ and $\text{Li}_3$. We have $$ \int_{0}^{1}\frac{\log(z)\log(1+z)}{1-z^2}\,dz = -\frac{\pi^2}{8}\log(2)+\frac{7}{16}\zeta(3)$$ and by Feynman's trick $$ \int_{0}^{1}\frac{\log(z)\log(1+az)}{1-z^2}\,dz = \int_{0}^{a}\frac{\pi^2(3u-1)+24\,\text{Li}_2(-u)}{24(1-u^2)}\,du $$ which ensures a rather amazing cancellation.