, 2008

Solution:
$x^{3} + y^{3} + k(x^{2} + y^{2}) = y^{3} + z^{3} + k(y^{2} + z^{2}) \Rightarrow x^{2} + xz + z^{2} = -k(x + z)$ \:(1)
and $y^{3} + z^{3} + k(y^{2} + z^{2}) = z^{3} + x^{3} + k(z^{2} + x^{2}) \Rightarrow y^{2} + yx + x^{2} = -k(y + x)$ \:(2)

- From (1) $-(2) \Rightarrow x + y + z = -k$ \:(*)

- If $x + z = 0$, then from (1) $\Rightarrow x^{2} + xz + z^{2} = 0 \Rightarrow (x + z)^{2} = xz \Rightarrow xz = 0$
So $x = z = 0$, contradiction since $x \neq z$ and therefore (1) $\Rightarrow -k = \frac{x^{2} + xz + z^{2}}{x + z}$
Similarly we have: $-k = \frac{y^{2} + yx + x^{2}}{y + x}$.
So $\frac{x^{2} + xz + z^{2}}{x + z} = \frac{y^{2} + x y + x^{2}}{x + y}$ from which $xy + yz + zx = 0$ \:(**)

We substitute $k$ in $x^{3} + y^{3} + k(x^{2} + y^{2}) = 2008$ from the relation () and using the (*), we finally obtain that $2 x y z = 2008$ and therefore $x y z = 1004$.