How to prove $(\frac{1}{5^3}-\frac{1}{7^3})+(\frac{1}{11^3}-\frac{1}{13^3})+(\frac{1}{17^3}-\frac{1}{19^3})+...=(1-\frac{\pi ^3}{18\sqrt{3}})$ How to prove 
$$ \sum_{k=1}^\infty \left[\frac{1}{(6k-1)^3} - \frac{1}{(6k+1)^3}\right] = 1 - \frac{\pi^3}{18\sqrt{3}}$$
I think this equality likes the Dirichlet Beta function.  The numerical value is checked but I don't have the proving.  Any help

Notice$\color{blue}{^{[1]}}$
$$\sum_{k=1}^\infty \left( \frac{1}{(6k-1)^3} - \frac{1}{(6k+1)^3}\right)
= \sum_{\substack{k=-\infty\\ k\ne 0}}^\infty \frac{1}{(6k-1)^3}
= 1 - \frac{1}{6^3}\sum_{k=-\infty}^\infty \frac{1}{(\frac16-k)^3}$$
Recall the infinite product expansion of $\sin x$
$$\sin x = x \prod_{k=1}^\infty \left( 1 - \frac{x^2}{k^2\pi^2}\right)$$
If one take logarithm and differentiate, one obtain an expansion of $\cot x$
$$\cot x = \sum_{k=-\infty}^\infty \frac{1}{x - k\pi}
\quad\iff\quad
\sum_{k=-\infty}^\infty \frac{1}{x-k} = \pi\cot(\pi x)\tag{*1}
$$
Differentiate the expansion on the right two more times, we get$\color{blue}{^{[2]}}$
$$\sum_{k=-\infty}^\infty \frac{1}{(x-k)^3} 
= \frac12 \frac{d^2}{dx^2} \left[ \sum_{k=-\infty}^\infty \frac{1}{x-k} \right]
= \frac{\pi}{2} \left[ \frac{d^2}{dx^2}\cot(\pi x) \right]
= \frac{\pi^3 \cos(\pi x)}{\sin(\pi x)^3}$$
As a result, the sum we want is
$$\sum_{k=1}^\infty \left( \frac{1}{(6k-1)^3} - \frac{1}{(6k+1)^3}\right)
= 1 - \frac{\pi^3}{6^3}\frac{\cos\frac{\pi}{6}}{\sin(\frac{\pi}{6})^3} \
= 1 - \frac{\pi^3}{18\sqrt{3}}
$$
Notes


*

*$\color{blue}{[1]}$ - Infinite sum of the form $\sum\limits_{k=-\infty}^\infty (\cdots)$ should be interpreted as $\lim\limits_{N\to\infty}\sum\limits_{k=-N}^N (\cdots)$.

*$\color{blue}{[2]}$ - To those who are not comfortable with the use of differentiation of 
an expansion. An alternate approach is start from the more well known expansion $(*1)$,
compute a contour integral of the form:
$$\frac{1}{2\pi i}\int_{|z|=R} \frac{\pi\cot(\pi z)}{(\frac16 - z)^3} dz$$
and show it vanishes as $R \to \infty$. 
The sum of the residues from $z \in \mathbb{Z}$  will be equal to the sum $\sum\limits_{k=-\infty}^\infty \frac{1}{\left(\frac16 - k\right)^3}$. It will be compensate by the residue
of the pole at $z = \frac16$. Since the pole at $z = \frac16$ is a triple one, its contribution will be proportional to $\frac{d^2}{dx^2}\pi\cot(\pi x)$.

