Cubefree sum

The Euler Product is $\prod_{p}\left(1+\frac{1}{p^2}+\frac{1}{p^4}\right)=\prod_{p}\frac{1-p^{-6}}{1-p^{-2}}=\frac{\zeta(2)}{\zeta(6)}=\frac{\pi^2}{6}\frac{945}{\pi^6}=\frac{315}{2\pi^4}$ and the answer is $\boxed{321}$

Two small issues: The $\pi^B$ should be $\pi^C$ and $A,B$ need to be co-prime positive integers.

Every positive integer $n$ can be written uniquely in the form $n = p\cdot q^3$ with $p$ cube-free. This provides a one-to-one correspondence between $\mathbb N$ and $(\mathbb N_{cf}, \mathbb N)$ .

Therefore $\sum_{n \in \mathbb N} \frac 1{n^2} =\sum_{p\in \mathbb N_{cf}}\sum_{q\in\mathbb N}\frac 1{(pq^3)^2} \\ = \left(\sum_{p\in \mathbb N_{cf}}\frac 1{p^2}\right)\ \left(\sum_{q\in\mathbb N}\frac 1{q^6}\right);$

If the desired value is $X$ then clearly $\zeta(2) = X\cdot \zeta(6)\ \ \ \therefore\ \ \ X = \frac{\zeta(2)}{\zeta(6)} = \frac{\pi^2/6}{\pi^6/945} = \frac{315}{2\pi^4}.$