Yes. Fourier series can help.
It is equivalent to finding the limiting function of the Fourier series
$\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {(2n - 1)t} \right)}}{{{{(2n - 1)}^3}}}} $ .
Note that$\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {nt} \right)}}{n}} $ converges to $(\pi - t)/2$ for $0 < t < 2 \pi $ .
Observe that
${\mathop{\rm Im}\nolimits} Log(1 - {e^{i2t}}) = - 2\sum\limits_{n = 1}^\infty {\frac{{\sin (2nt)}}{{2n}}} $ .
We can deduce this from
$Log(1 - {z^2}) = - 2\sum\limits_{n = 1}^\infty {\frac{{{z^{2n}}}}{{2n}}} $.
Hence $\sum\limits_{n = 1}^\infty {\frac{{\sin (2nt)}}{{2n}}} = - \frac{1}{2}{\mathop{\rm Im}\nolimits} Log(1 - {e^{i2t}}) = \frac{1}{2}\frac{1}{2}(\pi - 2t) = \frac{1}{4}(\pi - 2t)$ for $ 0 < t < \pi $.
Therefore,
$\sum\limits_{n = 1}^\infty {\frac{{\sin ((2n - 1)t)}}{{2n - 1}}} = \sum\limits_{n = 1}^\infty {\frac{{\sin (nt)}}{n}} - \sum\limits_{n = 1}^\infty {\frac{{\sin (2nt)}}{{2n}}} = \frac{1}{2}(\pi - t) - \frac{1}{4}(\pi - 2t) = \frac{\pi }{4}$ for $0 < t < \pi$ .
We may integrate the above Fourier series term by term to give the integral of the function on the right:
$ - \sum\limits_{n = 1}^\infty {\frac{{\cos \left( {(2n - 1)t} \right)}}{{{{(2n - 1)}^2}}}} + \sum\limits_{n = 1}^\infty {\frac{1}{{{{(2n - 1)}^2}}}} = \frac{\pi }{4}t$
I.e.,
$\sum\limits_{n = 1}^\infty {\frac{{\cos \left( {(2n - 1)t} \right)}}{{{{(2n - 1)}^2}}}} = \sum\limits_{n = 1}^\infty {\frac{1}{{{{(2n - 1)}^2}}}} - \frac{\pi }{4}t = \frac{{{\pi ^2}}}{8} - \frac{\pi }{4}t$ .
Integrating again gives:
$\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {(2n - 1)t} \right)}}{{{{(2n - 1)}^3}}}} = \frac{{{\pi ^2}}}{8}t - \frac{\pi }{8}{t^2}$
Now for $x \ge 1$ , $ t = \pi/x \le \pi $.
Substituting this value of $t$ in the above equation gives:
$\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {(2n - 1){\textstyle{\pi \over x}}} \right)}}{{{{(2n - 1)}^3}}}} = \frac{{{\pi ^3}}}{8}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)$ ,
which is equivalent to your equation for $x \ge 1$.
Best Answer
That is a special case of a general identity proved here.
With a Fourier-analytic approach, we may notice that: $$ \sum_{n\geq 1}\frac{(-1)^{n+1} \sin((2n-1)x)}{(2n-1)^2} $$ is the Fourier sine series of a triangle wave, hence: $$ \sum_{n\geq 1}\frac{(-1)^n}{(2n-1)^3} = -\int_{0}^{\pi/2}\left(\frac{\pi x}{4}\right)\,dx = \color{red}{-\frac{\pi^3}{32}}.$$