This is a computation made in Titchmash's Introduction to Zeta functions. I was trying to reverse the computation. However, I kept missing factors.
Consider $\frac{1- xyz^2}{(1-z)(1-xz)(1-yz)(1-xyz)}$. This is the trick used to compute $\frac{\zeta(s)\zeta(s-a)\zeta(s-b)\zeta(s-a-b)}{\zeta(2s-a-b)}$ where $\zeta$ is the Riemann zeta function. I always miss $1-xyz^2$ in partial fractions.
$$\frac{1}{(1-z)(1-xz)}=\frac{1}{1-x}(\frac{1}{1-z}-\frac{x}{1-xz})=f$$
$$\frac{1}{(1-yz)(1-xyz)}=\frac{1}{1-x}(\frac{1}{1-yz}-\frac{x}{1-xyz})=g$$
So $(1-xyz^2)fg$ is the original expression. It suffices to consider $fg$. There will be 4 terms with quadratic denominators. Consider $(1-x)^2fg=(\frac{1}{1-z}-\frac{x}{1-xz})(\frac{1}{1-yz}-\frac{x}{1-xyz})$ instead.
There are only 2 terms containing $\frac{1}{1-z}$.
$$\frac{1}{(1-z)(1-yz)}=(\frac{1}{1-z}-\frac{y}{1-yz})\frac{1}{1-y}$$
And
$$-\frac{x}{(1-z)(1-xyz)}=-\frac{x}{1-xy}(\frac{1}{1-z}-\frac{xy}{1-xyz})$$
Now combine $\frac{1}{1-z}$ coefficients. $\frac{1}{1-y}-\frac{x}{1-xy}=\frac{1-x}{(1-y)(1-xy)}$.
However I do not see removal of $(1-xy z^2)$(i.e. $\frac{1}{1-xy z^2}$ did not show up anywhere in the computation.)
The final step in the book gives $\frac{1- xyz^2}{(1-z)(1-xz)(1-yz)(1-xyz)}=\frac{1}{(1-x)(1-y)}(\frac{1}{1-z}-\frac{x}{1-xz}-\frac{y}{1-yz}+\frac{xy}{1-xyz})$.(If one starts with partial fraction between $\frac{1}{(1-z)(1-xz)}$ and $\frac{1}{(1-yz)(1-xyz)}$, then the desired result can be achieved.
$\textbf{Q:}$ Where is the mistake in above computation?(Say I only care to trace error from $\frac{1}{1-z}$ coefficients. The computation of those coefficients should be independent of each other.)