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How to prove that $$\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^32^k {2k\choose k}}=\frac1{4}\zeta(3)-\frac1{6}\ln^32?$$

A friend posted this nice problem on my FB group and I managed to evaluate it using the $\arcsin^2 x$ identity. I would like to see different approaches. Thanks.

My solution: Using the following identity: (see here) $$\arcsin^2z=\frac12\sum_{k=1}^\infty\frac{(2z)^{2k}}{k^2{2k \choose k}}$$

Set $\ z=\sqrt{\frac{x}{8}}$ then divide both sides by $x$ and integrate from $x=0$ to $-1$, to get \begin{align} S&=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^32^k {2k\choose k}}=-2\underbrace{\int_0^{-1}\frac{\arcsin^2\left(\sqrt{\frac x8}\right)}{x}\ dx}_{\large\arcsin\left(\sqrt{\frac x8}\right)=y}\\ &=-4\int_0^{\frac{\ln2}{2}i} y^2\cot y\ dy\overset{y=ix}{=}4\int_0^{\frac{\ln2}{2}} x^2\coth x\ dx \end{align} Lets find the antiderivative of the integral: \begin{align} I&=\int x^2\coth x\ dx\overset{IBP}{=}x^2\ln(\text{arcsinh}(x))-2\int x\ln(\text{arcsinh}(x))\ dx\\ &=x^2\ln(\text{arcsinh}(x))-2\int x\left\{x-\ln2-\ln(1-e^{-2x})\right\}\ dx\\ &=x^2\ln(\text{arcsinh}(x))-\frac23x^3+\ln2\ x^2-2\int x\ln(1-e^{-2x})\ dx\\ &=x^2\ln(\text{arcsinh}(x))-\frac23x^3+\ln2\ x^2+2\sum_{n=1}^\infty\frac1n\int xe^{-2nx}\ dx\\ &=x^2\ln(\text{arcsinh}(x))-\frac23x^3+\ln2\ x^2+2\sum_{n=1}^\infty\frac1n\left(-\frac{e^{-2nx}}{4n^2}-\frac{xe^{-2nx}}{2n}\right)\\ &=x^2\ln(\text{arcsinh}(x))-\frac23x^3+\ln2\ x^2-\frac12\sum_{n=1}^\infty\frac{(e^{-2x})^n}{n^3}-x\sum_{n=1}^\infty\frac{(e^{-2x})^n}{n^2}\\ &=x^2\left\{\ln x-\ln2-\ln(1-e^{-2x})\right\}-\frac23x^3+\ln2\ x^2-\frac12\operatorname{Li}_3(e^{-2x})-x\operatorname{Li}_2(e^{-2x})\\ &=\frac{x^3}{3}+x^2\ln(1-e^{-2x})-\frac12\operatorname{Li}_3(e^{-2x})-x\operatorname{Li}_2(e^{-2x})\\ \end{align}

Thus \begin{align} S&=4\left[\frac{x^3}{3}+x^2\ln(1-e^{-2x})-\frac12\operatorname{Li}_3(e^{-2x})-x\operatorname{Li}_2(e^{-2x})\right]_0^{\frac{\ln2}{2}}\\ &=4\left[\frac12\zeta(3)-\frac5{24}\ln^32-\frac12\operatorname{Li}_3\left(\frac12\right)-\frac{\ln2}{2}\operatorname{Li}_2\left(\frac12\right)\right]\\ &=4\left[\frac1{16}\zeta(3)-\frac1{24}\ln^32\right]\\ &\boxed{=\frac1{4}\zeta(3)-\frac1{6}\ln^32} \end{align}

Note that we used $\operatorname{Li}_2\left(\frac12\right)=\frac12\zeta(2)-\frac12\ln^22$ and $\operatorname{Li}_3\left(\frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$