Sum : $\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3}$ Prove that : $$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^3}=\frac{\pi^3}{32}.$$

I think this is known (see here), I appreciate any hint or link for the solution (or the full solution).

Splitting the series into 2 yields
$$
\begin{aligned}
\sum_{n=0}^{\infty} \frac{(-1)^{2}}{(2 n+1)^{3}} &=\sum_{n=0}^{\infty} \frac{1}{(4 n+1)^{3}}-\sum_{n=0}^{\infty} \frac{1}{(4 n+3)^{3}} \\
&=\sum_{n=0}^{\infty} \frac{1}{(4 n+1)^{3}}+\sum_{n=-1}^{\infty} \frac{1}{(-4 n-3)^{3}} \\
&=\frac{1}{64}\left[\sum_{n=0}^{\infty} \frac{1}{\left(n+\frac{1}{4}\right)^{3}}+\sum_{n=-1}^{-\infty} \frac{1}{\left(n+\frac{1}{4}\right)^{3}}\right] \\
&=\frac{1}{64} \sum_{n=-\infty}^{\infty} \frac{1}{\left(n+\frac{1}{4}\right)^{3}}
\end{aligned}
$$
Using the theorem $$\pi \cot (\pi z)=\sum_{n=-\infty}^{\infty} \frac{1}{n+z} ,$$
where $z \not\in \mathbb{Z}.$
Differentiating both sides w.r.t $z$ twice yields
$$
2 \pi^{3}\left(\cot ^{2} (\pi z)+1\right) \cot (\pi z)=\sum_{n=-\infty}^{\infty} \frac{2}{(n+z)^{3}}
$$
Putting $z=\frac{1}{4}$ yields
$$\boxed{\sum_{n=0}^{\infty} \frac{(-1)^{2}}{(2 n+1)^{3}} = \frac{1}{64} \cdot \frac{1}{2}\left[2 \pi^{3}\left(\cot ^{2}\left(\frac{\pi}{4}\right)+1\right) \cot \frac{\pi}{4}\right]=\frac{\pi^{3}}{32}}$$
