If for the real numbers $x, y,z, k$ the following conditions are valid, $x \ne y \ne z \ne x$ and
$x^3 +y^3 +k(x^2 +y^2) = y^3 +z^3 +k(y^2 +z^2) = z^3 +x^3 +k(z^2 +x^2) = 2008$, find the product $xyz$.
1. Given the equations:
   \[
   x^3 + y^3 + k(x^2 + y^2) = 2008
   \]
   \[
   y^3 + z^3 + k(y^2 + z^2) = 2008
   \]
   \[
   z^3 + x^3 + k(z^2 + x^2) = 2008
   \]

2. Subtract the first equation from the second:
   \[
   (y^3 + z^3 + k(y^2 + z^2)) - (x^3 + y^3 + k(x^2 + y^2)) = 0
   \]
   Simplifying, we get:
   \[
   z^3 - x^3 + k(z^2 - x^2) = 0
   \]
   Factorizing, we obtain:
   \[
   (z - x)(z^2 + zx + x^2 + k(z + x)) = 0
   \]
   Since \(x \ne z\), we have:
   \[
   z^2 + zx + x^2 + k(z + x) = 0
   \]

3. Similarly, subtract the second equation from the third:
   \[
   (z^3 + x^3 + k(z^2 + x^2)) - (y^3 + z^3 + k(y^2 + z^2)) = 0
   \]
   Simplifying, we get:
   \[
   x^3 - y^3 + k(x^2 - y^2) = 0
   \]
   Factorizing, we obtain:
   \[
   (x - y)(x^2 + xy + y^2 + k(x + y)) = 0
   \]
   Since \(x \ne y\), we have:
   \[
   x^2 + xy + y^2 + k(x + y) = 0
   \]

4. Similarly, subtract the third equation from the first:
   \[
   (x^3 + y^3 + k(x^2 + y^2)) - (z^3 + x^3 + k(z^2 + x^2)) = 0
   \]
   Simplifying, we get:
   \[
   y^3 - z^3 + k(y^2 - z^2) = 0
   \]
   Factorizing, we obtain:
   \[
   (y - z)(y^2 + yz + z^2 + k(y + z)) = 0
   \]
   Since \(y \ne z\), we have:
   \[
   y^2 + yz + z^2 + k(y + z) = 0
   \]

5. We now have the system of equations:
   \[
   x^2 + xy + y^2 + k(x + y) = 0
   \]
   \[
   y^2 + yz + z^2 + k(y + z) = 0
   \]
   \[
   z^2 + zx + x^2 + k(z + x) = 0
   \]

6. Subtract the first equation from the second:
   \[
   (y^2 + yz + z^2 + k(y + z)) - (x^2 + xy + y^2 + k(x + y)) = 0
   \]
   Simplifying, we get:
   \[
   z^2 - x^2 + yz - xy + k(z - x) = 0
   \]
   Factorizing, we obtain:
   \[
   (z - x)(z + x + y + k) = 0
   \]
   Since \(z \ne x\), we have:
   \[
   z + x + y + k = 0
   \]
   Thus:
   \[
   x + y + z = -k
   \]

7. Using the identity for the sum of cubes:
   \[
   x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)
   \]
   Given \(x + y + z = -k\) and substituting into the identity:
   \[
   x^3 + y^3 + z^3 - 3xyz = -k(x^2 + y^2 + z^2 - xy - yz - zx)
   \]

8. From the given conditions, we know:
   \[
   x^3 + y^3 + z^3 + k(x^2 + y^2 + z^2) = 2008
   \]
   Adding all three main equations:
   \[
   2(x^3 + y^3 + z^3) + 2k(x^2 + y^2 + z^2) = 6024
   \]
   Simplifying, we get:
   \[
   x^3 + y^3 + z^3 + k(x^2 + y^2 + z^2) = 3012
   \]

9. Using the identity and substituting \(x + y + z = -k\):
   \[
   x^3 + y^3 + z^3 - 3xyz = -k(x^2 + y^2 + z^2 - xy - yz - zx)
   \]
   Given \(xy + yz + zx = 0\), we have:
   \[
   x^2 + y^2 + z^2 = k^2
   \]
   Substituting back, we get:
   \[
   x^3 + y^3 + z^3 - 3xyz = -k(k^2)
   \]
   \[
   x^3 + y^3 + z^3 - 3xyz = -k^3
   \]
   Given \(x^3 + y^3 + z^3 + k^3 = 3012\), we have:
   \[
   -k^3 - 3xyz + k^3 = 3012
   \]
   \[
   -3xyz = 3012
   \]
   \[
   xyz = -\frac{3012}{3}
   \]
   \[
   xyz = -1004
   \]

The final answer is \(\boxed{-1004}\)