Riemann Zeta function

We have $\displaystyle\prod_{k=1}^\infty (1+\frac{1}{p_{k}^2}+\frac{1}{p_{k}^4})=\displaystyle\prod_{k=1}^\infty (\frac{p_{k}^4+p_{k}^2+1}{p_{k}^4})=\displaystyle\prod_{k=1}^\infty (\frac{p_{k}^4+2p_{k}^2+1-p_{k}^2}{p_{k}^4})=\displaystyle\prod_{k=1}^\infty \frac{(p_{k}^2-p_{k}+1)(p_{k}^2+p_{k}+1)}{p_{k}^4}= \displaystyle\prod_{k=1}^\infty \frac{(p_{k}+1)(p_{k}^2-p_{k}+1)(p_{k}-1)(p_{k}^2+p_{k}+1)}{p_{k}^4(p_{k}+1)(p_{k}-1)}=\displaystyle\prod_{k=1}^\infty \frac{p_{k}^6-1}{p_{k}^6-p_{k}^4} =\displaystyle\prod_{k=1}^\infty \frac{1-\frac{1}{p_{k}^6}}{1-\frac{1}{p_{k}^2}}=\displaystyle\prod_{k=1}^\infty \frac{(1-\frac{1}{p_{k}^2})^{-1}}{(1-\frac{1}{p_{k}^6})^{-1}}=\frac{\zeta {(2)}}{\zeta {(6)}}=\frac{315}{2\pi^4}$

First:

$(1+a^2+a^4)(1-a^2)=(1-a^2)+(a^2-a^4)+(a^4-a^6)=1-a^6\\1+a+a^2=\frac{1-a^6}{1-a^2}$

Second:

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

Third:

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

Finally:

$\prod_{p\in prime}\left(1+\frac{1}{p_k}+\frac{1}{p_k^2}\right)=\prod_{p\in prime}\left(\frac{1-\frac{1}{p_k^6}}{1-\frac{1}{1-p_k^2}}\right) =\frac{\prod_{p\in prime}\left(1-\frac{1}{p_k^6}\right)}{\prod_{p\in prime}\left(1-\frac{1}{p_k^2}\right)} \\=\frac{\prod_{p\in prime}\left(1-\frac{1}{p_k^2}\right)^{-1}}{\prod_{p\in prime}\left(1-\frac{1}{p_k^6}\right)^{-1}} =\frac{\prod_{p\in prime}\left(\frac{1}{1-p^{-2}}\right)}{\prod_{p\in prime}\left(\frac{1}{1-p^{-6}}\right)} =\frac{\zeta(2)}{\zeta(6)}=\frac{315}{2\pi^4}$