Finding a closed form for $\sum_{k=1}^{\infty}\frac{1}{(2k)^5-(2k)^3}$

$$  \frac{1}{(2k)^5 - (2k)^3} + \frac{1}{(2k)^3} = \frac{1 + (2k)^2 - 1}{(2k)^5 - (2k)^3} = \frac{1}{(2k)^3 -2k}$$

So by Ramanujan's result:

$$\sum_{k=1}^{\infty} \frac{1}{(2k)^5 - (2k)^3} = \ln(2) - \frac{1}{2} - \frac{1}{8}\zeta(3)$$