Use Fourier series of $f(x)=x(\pi-|x|)$ in $(-\pi,\pi)$ to compute the series $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{(2n-1)^3}.$ The fourier series I obtained was
$$x(\pi-|x|)=\frac{4}{\pi}\sum_{n\in\mathbb{N}}\frac{(-1)^{n+1}+1}{n^3}\sin(nx),\tag1$$
which I checked is correct by plotting RHS and LHS. I'm not sure what to set $x$ as when evaluating the RHS, since I often get $0$ due to the sine.
I'm also having trouble converting the denominator in the sum fom $n^3$ to $(2n-1)^3$.
Some have noted that my Fourier expansion is wrong. I'm not sure about its correctness but based on the plots on Maple, it seems correct. The blue is the original function and the red is the Fourier series.

 A: Now that you got the right expansion, it is not difficult at all. Choose $x=\pi/2$ which leads to the following
\begin{align*}
\frac\pi2\left(\pi-\left|\frac\pi2\right|\right)&=\frac4\pi\sum_{n\ge 1}\frac{(-1)^{n+1}+1}{n^3}\sin\left(n\frac\pi2\right)\\
\frac{\pi^3}{16}&=\sum_{n\ge 1}\frac{(-1)^{n+1}+1}{n^3}\sin\left(n\frac\pi2\right)
\end{align*}
Now we can distinguish between even and odd $n$. For even $n$ the term $\sin(n\pi/2)$ will be $0$ and therefore these sum terms cancel, whereas for odd $n$ the value will alternate between $-1$ and $1$ (and equal $(-1)^n$ for odd and even $n$ in the new index). So we can further conclude that
\begin{align*}
\frac{\pi^3}{16}&=\sum_{n\ge 1}\frac{(-1)^{n+1}+1}{n^3}\sin\left(n\frac\pi2\right)\\
&=\sum_{n\ge0}\frac{(-1)^{(2n+1)+1}+1}{(2n+1)^3}(-1)^n\\
&=2\sum_{n\ge 1}\frac{(-1)^{n+1}}{(2n-1)^3}
\end{align*}

$$\therefore~\sum_{n\ge 1}\frac{(-1)^{n+1}}{(2n-1)^3}~=~\frac{\pi^3}{32}$$

Note that the here-occuring sum is a particular case of the Dirichlet Beta Function $\beta(s)$; precisely we just managed to show that $\beta(3)={\pi^3}/{32}$.
A: A hint to get you started - 
$$ \sum_{n=1}^\infty ((-1)^{n+1} +1 )a_n =2a_1 + 0 + 2a_3+0 + \dots = 2 \sum_{k=1}^\infty a_{2k-1}  $$
and then the choice $x=\pi/2$ should allow you to conclude.
