Consider the series
$$\frac34\sum_{n\ge1}\left[\frac{(-1)^n}{(2n)^3}-\frac2{n^3}\right]=\frac3{16}\sum_{n\ge1}\frac{(-1)^n-8}{n^3}$$
I have come across this sum while evaluating the integral
$$\int_{0}^{\pi/4}\frac{x^3}{\sin^2x}\,{\rm d}x=\frac{3\pi}{4}G-\frac{\pi^3}{64}+\frac{3\pi^2}{32}\log2-\frac{105}{64}\zeta(3)$$
where $G$ denotes Catalan's Constant and $\zeta(z)$ the Riemann Zeta Function. To do so I followed this approach (Solution to Problem $18$ given by ysharifi)
Integration by parts with $x^3=u$ and $\frac{{\rm d}x}{\sin^2 x}={\rm d}v$ reduces the problem to $\int_0^{\pi/4}x^2\cot x\,{\rm d}x$. Again, integration by parts with $x^2=u$ and $\cot x\,{\rm d}x={\rm d}v$ reduces the problem to $\int_0^{\pi/4}x\ln \sin x\,{\rm d}x$ and this one can by easily found using the identity $$\ln \sin x =-\ln 2-\sum_{n=1}^{\infty}\frac{\cos(2nx)}{n}$$ which holds for $x\in(0,\pi)$
The sum I failed to express in terms of the Riemann Zeta Function occured within the last step of calculation. To put in a nutshell I do not know how to show
$$\frac3{16}\sum_{n\ge1}\frac{(-1)^n-8}{n^3}=-\frac{105}{64}\sum_{n\ge1}\frac1{n^3}=-\frac{105}{64}\zeta(3)$$
As this is the last step of evaluating the integral I would be glad if someone could explain to me how to show the equalitiy of these two sums.
Thanks in advance!
$$ \begin{align} \color{red}{S} & =+\frac{3}{16}\sum_{n=1}^{\infty}\frac{(-1)^n-8}{n^3} =-\frac{3}{16}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^3}-\frac{3}{16}\sum_{n=1}^{\infty}\frac{8}{n^3} \\[2mm] & =-\frac{3}{16}\eta(3)-\frac{3}{2}\zeta(3) =\left(-\frac{3}{16}\left(1-2^{1-3}\right)-\frac{3}{2}\right)\zeta(3) =\color{red}{-\frac{105}{64}\zeta(3)} \end{align} $$