Let $1 + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{7^3} + \dots=s$, show that then $\sum_1^\infty\frac{1}{n^3}=\frac{8}{7}s$ Let $1 + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{7^3} + \dots=s$, show that then $\sum_1^\infty\frac{1}{n^3}=\frac{8}{7}s$.
This is the last part of a problem that I am working on. So far, we have shown that the cosine series for $x^2$ is 
$x^2=\frac{\pi^2}{3}+4\sum_1^\infty (-1)^n\frac{\cos(nx)}{n^2}$
The sine series for $x^2$ is
$x^2=2\pi\sum_1^\infty(-1)^{n+1}\frac{\sin nx}{n}-\frac{8}{\pi}\sum_1^\infty\frac{\sin ((2n-1)x)}{(2n-1)^3}$
The sine series for x is 
$x=2(\sin x -\frac{\sin 2x}{2}+ \frac{\sin 3x}{3}-\dots)$
and using the above we got  that 
$\frac{\pi^3}{32}=1 - \frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+\dots$.
Not sure if any of this matters, I just don't know how to use this information to solve the last part. I don't really know if this information is needed at all, but I assume it is. 

Let $\sum_{n=1}^\infty \frac{1}{(2n-1)^3}=s$, then we have
\begin{eqnarray*}
\sum_{n=1}^\infty \frac{1}{n^3} &=& \sum_{n=1}^\infty \sum_{m=0}^\infty \frac{1}{(2n-1)^3} \frac{1}{2^{3m}} \\
&=& \sum_{m=0}^\infty \frac{1}{8^m} \sum_{n=1}^\infty  \frac{1}{(2n-1)^3} \\
&=& \frac{1}{1-\frac{1}{8}} s \\
&=& \frac{8}{7} s
\end{eqnarray*}
Where we first count all odd numbers, then the numbers divisble by $2$ but not $4$, then divisble by $4$ not $8$ et cetera. This gives us the sum over $m$ in the first line.
And we may switch order of summation as we sum only positive numbers.
