I am asked to solve the following problem
$$\sum_{n=0}^\infty{\frac{\sin^3{(3^n)}}{3^n}}$$
My work:
Using the identity $\sin{(3x)}=3\sin x-4\sin^3x$
$$\sum_{n=0}^\infty{\frac{\sin^3{(3^n)}}{3^n}}=\frac{1}{4}\sum_{n=0}^{\infty}{\frac{3\sin 3^n-\sin{3^{n+1}}}{3^n}}$$
From here I do not know how to procede, even tough it looks like it is a telescoping series, I cannot find a way to arrive at it.
I also tried:
$$n\rightarrow3^n\\ \sum_{n=1}^{\infty}{\frac{\sin^3{n}}{n}}\\ \text{Applying the identity}\\ \frac{1}{4}\sum_{n=1}^{\infty}{\frac{3\sin n-\sin{n}}{n}}$$
Which again looks like it might telescope, but I do not see how to arrive at the result.
Any help, hint or solution is welcome!