Given the following function defined in $\mathbb{R}^3$, with the restriction $0\leq p\leq 1$:
$$f(x,y,z)=p^{| x+y-z| +| x-y+z| +| -x+y+z| +| x+y+z| +| x-y| +| x+y| +| x-z| +| x+z| +| x| +| y-z| +| y+z| +| y| +| z| }$$
I need to compute the discrete sum over its domain as follows:
$$\sum_{x=-\infty}^{\infty} \sum_{y=-\infty}^{\infty} \sum_{z=-\infty}^{\infty} f(x, y, z)$$
For this purpose, I tried to use the same methodology as in the following case:
$$\sum _{x=-\infty }^{\infty } \sum _{y=-\infty }^{\infty } p^{| y-x| +| x+y| +| x| +| y| }=8\sum _{x=2}^{\infty } \sum _{y=1}^{x-1} p^{x+x+x+y+y-y}+4\sum _{x=1}^{\infty } p^{3 x}+4\sum _{x=1}^{\infty } p^{4 x}+1=\frac{p \left(p^5-p^4+p^3+2 p^2+p-1\right)+1}{(p-1)^2 \left(p^2+1\right) \left(p^2+p+1\right)}$$
As seen, a similar function defined in $\mathbb{R}^2$ can be summed over its domain by splitting the sum into smaller sectors concerning its symmetry axes, in addition to the axes themselves. However, I can't manage to visualize the 3D sectors $f(x,y,z)$ must be divided into. So, is there any way to work out this sum algebraically through a process, specific to each dimension, and dependent on the symmetry axes defined on the exponent of $f(x,y,z)$?