If for the real numbers $x, y, z, k$ the following conditions are valid, $x \neq y \neq z \neq x$ and $x^{3}+y^{3}+k\left(x^{2}+y^{2}\right)=y^{3}+z^{3}+k\left(y^{2}+z^{2}\right)=z^{3}+x^{3}+k\left(z^{2}+x^{2}\right)=2008$, find the product $x y z$.
$x^{3}+y^{3}+k\left(x^{2}+y^{2}\right)=y^{3}+z^{3}+k\left(y^{2}+z^{2}\right) \Rightarrow x^{2}+x z+z^{2}=-k(x+z):(1)$ and $y^{3}+z^{3}+k\left(y^{2}+z^{2}\right)=z^{3}+x^{3}+k\left(z^{2}+x^{2}\right) \Rightarrow y^{2}+y x+x^{2}=-k(y+x):(2)$

- From (1) $-(2) \Rightarrow x+y+z=-k:(*)$
- If $x+z=0$, then from $(1) \Rightarrow x^{2}+x z+z^{2}=0 \Rightarrow(x+z)^{2}=x z \Rightarrow x z=0$

So $x=z=0$, contradiction since $x \neq z$ and therefore $(1) \Rightarrow-k=\frac{x^{2}+x z+z^{2}}{x+z}$

Similarly we have: $-k=\frac{y^{2}+y x+x^{2}}{y+x}$.

So $\frac{x^{2}+x z+z^{2}}{x+z}=\frac{y^{2}+x y+x^{2}}{x+y}$ from which $x y+y z+z x=0:(* *)$.

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

Remark: $x, y, z$ must be the distinct real solutions of the equation $t^{3}+k t^{2}-1004=0$. Such solutions exist if (and only if) $k>3 \sqrt[3]{251}$.