Show that $\prod_{p} (1 - \frac{1}{p^{2s}}(1-\frac{1}{p^s})^{-2})$ is a Dirichlet series

Question

As a part of the proof that $$\sum_{n \leq x} \mu (n)d(n) = \mathcal{O}(x e^{-c(\log x)^{1/10}}),$$ we first write the sum as $$\sum_{n} \frac{\mu (n)d(n)}{n^s} = \prod_{p}(1-\frac{2}{p^s})$$ which is equivalent to $$\prod_{p} (1 - \frac{1}{p^s})^2(1 - \frac{1}{p^{2s}}(1-\frac{1}{p^s})^{-2}).$$ I need to show that the above product can be written as $\frac{g(s)}{\zeta^2 (s)}$ where $g(s)$ is a Dirichlet series absolutely convergent for $\Re(s) > 1/2$. How can I show that $$\prod_{p} (1 - \frac{1}{p^{2s}}(1-\frac{1}{p^s})^{-2})$$ is equivalent to $g(s)$? Probably Euler product can solve this problem, but I am not sure how.

Answer 1

By the Euler product formula, $$\zeta(s) = \prod_{p}{\left(1 - \frac{1}{p^s}\right)}^{-1}$$

which means

$$\prod_{p}{\left(1 - \frac{1}{p^s}\right)} = \frac{1}{\zeta(s)}.$$

Then taking your equation

$$\prod_p{\left(1 - \frac{1}{p^s}\right)^2 \left(1 - \frac{1}{p^{2s}}\left(1-\frac{1}{p^s}\right)^{-2}\right)} = \frac{g(s)}{\zeta^2(s)},$$

we see that

$$\frac{1}{\zeta^2(s)} \cdot \prod_p{\left(1 - \frac{1}{p^{2s}}\left(1-\frac{1}{p^s}\right)^{-2}\right)} = \frac{g(s)}{\zeta^2(s)}.$$

And from here you can deduce your answer.