8. The number of all integer solutions $(x, y, z)$ for the equation $x y z=2009$ is $\qquad$ groups.
8. 72 .

First consider the integer solutions for $00$, it is easy to know that there are
$$
2 \mathrm{~A}_{3}^{1}+2 \mathrm{~A}_{3}^{3}=18
$$

sets of positive integer solutions.
For each set of positive integer solutions, adding two negative signs can yield 3 sets of integer solutions with two negatives and one positive.

Therefore, the total number of integer solutions that satisfy the condition is
$$
18 \times(1+3)=72 \text { (sets). }
$$