Is the following sum irrational?

$$S = \displaystyle \sum_{n \text{ squarefree}, n \geq 1} \frac{1}{n^3}$$

The sum clearly converges, so it is bounded above by $\zeta(3) = \displaystyle \sum_{n \geq 1} \frac{1}{n^3}$. In fact, since we have $$\displaystyle \sum_{n \text{ squarefree}, n \geq 1} \frac{1}{n^3} = \prod_p \left(1 + \frac{1}{p^3}\right) = \sum_{n=1}^\infty \frac{\mu^2(n)}{n^3},$$ the sum is exactly equal to $\displaystyle \frac{\zeta(3)}{\zeta(6)}$.

Hence if $S$ is irrational, then it would show that $\zeta(3)$ is not a rational multiple of $\pi^6$. Since we know that $\zeta(3)$ is not rational given Apery's work, this is a slight strengthening of the result of Apery.

It seems that Apery's approach should still work for $S$, but I am not sure. Does anyone know the answer or the plausibility of Apery's approach working?