, 2008

Solution:
The last equation implies $x y z = 2008(x y + y z + z x)$, therefore
$$x y z - 2008(x y + y z + z x) + 2008^{2}(x + y + z) - 2008^{3} = 0.$$
$(x - 2008)(y - 2008)(z - 2008) = 0$.
Thus one of the variables equals $2008$. Let this be $x$. Then the first equation implies $y = -z$. From the second one it now follows that
$$2 y^{2} = 6024^{2} - 2008^{2} = 2008^{2}(9 - 1) = 2 \cdot 4016^{2}.$$
Thus $(x, y, z)$ is the triple $(2008, 4016, -4016)$ or any of its rearrangements.