Set $a=2008$ then $xy+yz+xz=-4a^2, xyz=-4a^3$ and so $(x,y,z)$ -roots of $X^3-aX^2-4a^2X+4a^3=0$ If we set $X=aY$ then we get $Y^3-Y^2-4Y+4=0$ with roots $Y=1,2,-2$ and so $(x,y,z)=(2008,4016,-4016)$

RagvaloD wrote: Set $a=2008$ then $xy+yz+xz=-4a^2, xyz=-4a^3$ and so $(x,y,z)$ -roots of $X^3-aX^2-4a^2X+4a^3=0$ If we set $X=aY$ then we get $Y^3-Y^2-4Y+4=0$ with roots $Y=1,2,-2$ and so $(x,y,z)=(2008,4016,-4016)$ It's wrong! \begin{align*}\\ \ 3(x^2+y^2+z^2)-(x+y+z)^2=(x-y)^2+(y-z)^2+(z-x)^2=0 \end{align*}

Hint: Observe that $xyz-2008(xy+yz+zx)+2008^2(x+y+z)-2008^3=0$, but $xyz-2008(xy+yz+zx)+2008^2(x+y+z)-2008^3=(x-2008)(y-2008)(z-2008)$. Thus one of the variable is 2008.