Prove that $\sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right)=\left(\frac{1}{64}+\frac{3}{128\sqrt{2}}\right)\pi^3$ Prove that $$\sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right)=\left(\frac{1}{64}+\frac{3}{128\sqrt{2}}\right)\pi^3$$
I don't have an idea about how to start.

This answer uses the hint using the polygamma function
$$
\psi^{(2)}(z) = -\int_0^1 \frac{t^{z-1}}{1-t}\ln^2t dt.
$$
First, using the expansion of $(1-t)^{-1}$, the polygamma function $\psi^{(2)}(x)$ can be written as
\begin{align}
\psi^{(2)}(z) &= -\sum_{n=0}^\infty \int_0^1 t^{n+z-1}\ln^2 tdt \cr
 &= -\sum_{n=0}^\infty \int_0^\infty s^2 e^{-(n+z)s}ds
\qquad \qquad \qquad (t=e^{-s}) \cr
&= -2\sum_{n=0}^\infty \frac{1}{(n+z)^3}.
\end{align}
Therefore, 
$$
\sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right) = \frac{1}{1024}\left(\psi^{(2)}\left(\frac{7}{8}\right)
-\psi^{(2)}\left(\frac{1}{8}\right)\right).
$$
Using the reflection relation
$$
\psi^{(2)}(1-z)-\psi^{(2)}(z) = \pi\frac{d^2}{dz^2}\cot \pi z
$$
with $z=1/8$,  the sum can be written as
$$
\left.\frac{\pi}{1024}\frac{d^2}{dz^2}\cot \pi z\right|_{z=1/8}
=\frac{\pi^3}{512}\left(
\cot\frac{\pi}{8}+\cot^3\frac{\pi}{8}\right).
$$
Finally, using trigonometric identities (half angle),we can get $\cot \pi/8 = 1+\sqrt{2}$, and 
$$
\sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right)
=\frac{\pi^3}{512}\Big(
1+\sqrt{2}+(1+\sqrt{2})^3
\Big)=\frac{\pi^3}{256}(4+3\sqrt{2}).
$$
