As in the title, I want to find the Dirichlet series $F$ of the indicator function for cube full integers $f(n)=1 \iff p^3|n, \forall p|n$ and $f(n)=0$ otherwise. Since $f$ is clearly multiplicative, $F$ could be expressed with Euler product:
$$F(s)=\prod_{p}(1+\frac{f(p)}{p^s}+\frac{f(p^2)}{p^{2s}}+\frac{f(p^3)}{p^{3s}}...)=\prod_{p}(1+\frac{1}{p^{3s}}+\frac{1}{p^{4s}}...)$$
I know that the Dirichlet series $G$ of the indicator function for square full integers has the property
$$G(s)=\prod_{p}(1+\frac{g(p)}{p^s}+\frac{g(p^2)}{p^{2s}}+\frac{g(p^3)}{p^{3s}}...)=\prod_{p}(1+\frac{1}{p^{2s}}+\frac{1}{p^{3s}}...)=\frac{\zeta(2s)\zeta(3s)}{\zeta(6s)}$$
I can prove this, but I have no idea where the formula came from. I'm hoping that I could prove something similar for $F$. Any help is appreciated, thank you in advance.
Edit:Here is the proof (I'd much rather call it a verification) for $G$.
Let $x=p^s$. Then LHS is the product of:
$$1+\frac{1}{p^{2s}}+\frac{1}{p^{3s}}...=1+\frac{1}{x^2}+\frac{1}{x^3}...= \frac{x^2-x+1}{x^2-x}$$
$\zeta(2s)$, $\zeta(3s)$, $(\zeta(6s))^{-1}$ are the product of:
$$1+\frac{1}{p^{2s}}+\frac{1}{p^{4s}}...=1+\frac{1}{x^2}+\frac{1}{x^4}...=\frac{1}{1-\frac{1}{x^2}}$$
$$1+\frac{1}{p^{3s}}+\frac{1}{p^{6s}}...=1+\frac{1}{x^3}+\frac{1}{x^6}...=\frac{1}{1-\frac{1}{x^3}}$$
$$(1+\frac{1}{p^{6s}}+\frac{1}{p^{12s}})^{-1}...=(1+\frac{1}{x^6}+\frac{1}{x^{12}}...)^{-1}=1-\frac{1}{x^6}=\frac{x^6-1}{x^6}$$
Thus RHS is the product of
$$\frac{1}{1-\frac{1}{x^2}}\frac{1}{1-\frac{1}{x^3}}\frac{x^6-1}{x^6}=\frac{x^6-1}{x^6-x^4-x^3+x}$$
The proof is finished by noting that
$$(x^2-x+1)(x^6-x^4-x^3+x)=x^8-x^7-x^2+x=(x^2-x)(x^6-1)$$
I can't generalize this to prove something about $F$.
"I have no idea where the formula comes from":
As you noted, and using $x$ as you do, an Euler factor in the Euler product for $F$ is \[ 1+x^3+x^4+x^5+\cdot \cdot \cdot \hspace {10mm}(1)\] whereas a factor in the product for $\zeta (s)$ is \[ \frac {1}{1-x}.\] To write something as a "nice" product, what you want to do is write it as something involving these $1-x$ factors, or at least involving polynomial factors. That's where the formula "comes from".
So in this case you write (1) as \[ 1+x^3(1+x+x^2+\cdot \cdot \cdot )=1+\frac {x^3}{1-x}=\frac {1-x+x^3}{1-x}\] and recognise the $1-x$ factor as that for $\zeta (s)$, so that \[ F(s)=\zeta (s)\prod _{p}(1-p^{-s}+p^{-3})\] and that's all you can say, because as reuns said you can't simplify anything in that product.