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.

For $b \in \{0,\ldots,7\}$, consider the function $G(b) := \sum_{n \geq 1} \frac{e^{-2\pi i nb/8}}{n^3}$. Observe that
\begin{equation*}
G(b) = \sum_{0 \leq a \leq 7} e^{-2\pi i ab/8} \sum_{n \geq 1} \frac{1}{(8n-a)^3}.
\end{equation*}
Now, we know that if $a,a' \in \{0,\ldots,7\}$ then
\begin{equation*}
\frac{1}{8}\sum_{0 \leq b \leq 7} e^{2\pi i(a-a')b/8} = \begin{cases} 1& \text{ if $a = a'$} \\ 0& \text{ otherwise}. \end{cases}
\end{equation*}
Therefore, we have
\begin{equation*}
\frac{1}{8} \sum_{0 \leq b \leq 7}e^{-2\pi i(7b)/8} G(b) = \sum_{n \geq 1} \frac{1}{(8n-7)^3},
\end{equation*}
and thus your sum in question is precisely 
\begin{equation*}
\frac{1}{8}\sum_{0 \leq b \leq 7} \left(e^{-2\pi i(7b)/8} -e^{-2\pi ib/8}\right)G(b).
\end{equation*}
Now, observe that $G(b) = \sum_{n \geq 1} a_n e^{-2\pi int}$, where $t = \frac{b}{8}$ and $a_n = \frac{1}{n^3}$. This looks like a Fourier series. Can you determine which function it represents? If so then you would be nearly done.
