Finding $\sum_{r=2}^{\infty}\ln\left(1-\frac{1}{r^3}\right)$ 
Find $$\sum_{r=2}^{\infty}\ln\left(1-\frac{1}{r^3}\right)$$

I solved the simpler versions of these:
$$\sum_{r=2}^{\infty}\ln\left(1-\frac{1}{r}\right)\;\;,\;\;\sum_{r=2}^{\infty}\ln\left(1-\frac{1}
{r^2}\right)$$
by factorizing which yields a telescoping sum. However the cube version certainly breaks this trend.
What else can be done here? Does a closed form exist?

Hint:
$$1-\dfrac1{r^3}=\dfrac{(r-1)(r^2+r+1)}{r^3}=\dfrac{f(r-1)g(r-1)}{f^3(r)}$$
where $g(r-1)=r^2+r+1\implies g(r)=(r+1)^2+(r+1)+1=r^2+3r+3$
and $f(r)=r$
Similarly,  $$1-\dfrac1{(r+1)^3}=\dfrac{r(r^2+3r+3)}{(r+1)^3}=\dfrac{f(r)g(r)}{f^3(r+1)}$$
$$\implies\prod_{r=2}^n\left(1-\dfrac1{r^3}\right)=\prod_{r=2}^ng(r)\cdot\dfrac{f(1)}{f^2(2)f^2(3)\cdots f^2(n)f^3(n+1)}=\dfrac{f(1)}{f(n+1)}\prod_{r=2}^n\dfrac{g(r)}{f^2(r)}$$
$\prod_{r=2}^nf(r)=n!$
