Find the volume of the region in space defined by $|x+y+z|+|x+y-z|\le 8$ where $x,y,z\ge 0$. 
Find the volume of the region in space defined by $|x+y+z|+|x+y-z|\le
 8$ where $x,y,z\ge 0$.

I have honestly no idea on how to approach this problem. 
By browsing a little bit on the internet I have found a similar question with $x+y+z < 4$ which turns out to be  two tetrahedrons with same base but I don't know how I can (if I can)turn my equation in this form.
Any help is appreciated.

Just use the definition of absoloute value: if $x,y,z\ge 0$ then 
$|x+y+z|+|x+y-z|\le 8$ amounts to 
$2x+2y\leq 8\Rightarrow x+y\leq 4\ $ if $x+y-z\geq 0$
and 
$2z\leq 8\Rightarrow z\leq 4\ $ if $x+y-z\leq 0$
so we can find the limits by integrating first with respect to $x:0\ $ to $4-y\ $ then project onto the $yz\ $plane, obtaining $y\leq 4\ $ whenever $y-z\geq 0\ $ and $z\leq 4\ $ if $y-z\leq 0.$ and this is just the square $[0,4]\times [0,4]$ so the integral is 
$\int_0^4\int_0^4\int_0^{4-y} \,dx\,dz\,dy.$
