Euler product on congruence classes

Question

Is it possible to evaluate the Euler product $\prod\limits_{p\equiv 1\pmod{6}} \frac{p^3}{p^3-1}$ in terms of $\zeta(3)$?

Thanks, in advance.

Answer 1

The reciprocal number $\prod_{p\equiv 1 \pmod 6}(1-\frac{1}{p^3})\approx 0.996401\ldots$ can perhaps be obtained from the two numbers $L(3,1) = \prod_{n\equiv (1,5) \pmod 6} (1-\frac{1}{p^3}) = 91\zeta(3)/108$ and $L(3,2) = \prod_{n \equiv 1\pmod 6}(1-\frac{1}{p^3})\prod_{n\equiv 5\pmod 6}(1+\frac{1}{p^3}) = \pi^3/(18\surd 3)$ shown in Table 22 of http://arxiv.org/abs/1008.2547 for $m=6$. The value $1.00361130\ldots=1/0.996401\ldots$ is also tabulated numerically in Section 3.3.