from first and second we get x^2+xz+z^2=-k(x+z) from second and third condition we get similar result after that substract one from another and get x+y+z=-k now it is obvious that x+z/=0 now we can equalize k's from condition and result to get xy+yz+zx=0 and finaly replace k -(x+y+z) and manipulate a little and get xyz=1004 so we are done

Add the first three expressions (all equal to $2008$) to get $x^3+ y^3+ z^3+ k(x^2+ y^2+ z^2) = 3012$ Now subtract each of the first three expressions and we have $x^3- kx^2- 1004 = 0$ $y^3- ky^2- 1004 = 0$ $z^3- kz^2- 1004 = 0$ Since $x \ne y \ne z \ne x$ then $x, y, z$ are the three (distinct) roots of $x^3- kx^2- 1004 = 0$. $xyz$ is the product of roots, which by Vieta's we get as $-\frac{-1004}{1} = \boxed{1004}.$

Subtracting the first two equations we get: $(x-z)(x^2+xz+z^2+kx+kz)=0$ Since $x \ne z$ we get that $x^2+xz+z^2+k(x+z)=0$. Similarly we get : $x^2+xy+y^2+k(x+y)=x^2+xz+z^2+k(x+z)=y^2+yz+z^2+k(y+z)=0$ $(*)$ Subtracting the first two we get : $y^2-z^2+x(y-z)+k(y-z)$ $\iff (y-z)(x+k)+(y-z)(y+z)=0$ Dividing through by $(y-z)$ ($y\ne z$) we get that $x+y+z=-k$ We know $(y+z)(x^2+xy+y^2)=(x+y)(y^2+yz+z^2)$ by expressing $k$ from two different equations $(*)$ $\iff xy+yz+zx=0$ After finding that $x+y+z=-k$ , $xy+yz+zx=0$ we get $x^2+y^2+z^2=k^2$. By adding the three main equations we get: $2(x^3+y^3+z^3)+2k(x^2+y^2+z^2)=6024$ $\iff x^3+y^3+z^3+k^3=3012$ (from the conditions above) $\iff 3xyz+(x+y+z)(x^2+y^2+z^2-xy-yz-zx)+k^3=3012$ $\iff 3xyz-k(k^2-0)+k^3=3012$ $\iff \boxed{xyz=1004}$

From first and second we get $$(x-z)(x^2+xz+z^2+kx+kx)=0,...$$But from $x \ne y \ne z \ne x$ we get: $$1) x^2+xz+z^2+kx+kz=0$$$$2) y^2+yz+z^2+ky+kz=0$$$$3) x^2+xy+y^2+kx+ky=0$$When we subtract first and second new equation we get: $$(x-y)(x+y+z+k)=0$$$$ x+y+z=-k$$Now,when we add all three equations,we have: $$ 2(x+y+z)^2 - 3(xy+yz+yz) + 2k(x+y+z) = 0$$$$ 2k^2-3(xy+yz+xz) +2k^2 = 0 $$$$ xy+yz+xz=0$$From $$xy+yz+xz=0$$and $$x+y+z=-k $$we get $$x^2 + y^2 + z^2= k^2$$$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)=-k^3$$and $$(x+y+z)^3=-k^3$$then we have: $$x^3+y^3+z^3+3xy^2+3x^2y+3yz^2+3y^2z+3z^2x+3z^2y+6xyz=x^3+y^3+z^3-3xyz$$$$3xy^2+3x^2y+3yz^2+3y^2z+3z^2x+3z^2y+9xyz=0 ***$$From the condition:$$x^3+y^3-(x+y+z)(x^2+y^2)=2008$$we get: $$x^2y+x^2z+y^2x+y^2z=-2008.$$Similarly we get: $$y^2z+y^2x+z^2x+z^2y=-2008.$$$$x^2y+x^2z+z^2x+z^2y=-2008.$$Then we have: $$2*(x^2y+x^2z+y^2x+y^2z+z^2x+z^2y)=-2008*3$$, $$x^2y+x^2z+y^2x+y^2z+z^2x+z^2y=-1004*3$$From *** we have: $$ 9*-1004+9xyz=0$$then $$xyz=1004$$

Similar, from first two equations we get $x^3+kx^2+ky^2=y^3+z^3+ky^2+kz^2$ and after factorization $x^2+xz+z^2=-k(x+z)$. Similar, $y^2+xy+x^2=-k(x+y)$ We will subtract these equations to get $x+y+z=-k$. From first two equations we have that $-k=\frac{x^2+xz+z^2}{x+z}=\frac{x^2+xy+y^2}{x+y}$. So after factorization $x+y-\frac{zx}{x+z}=x+y-\frac{xy}{x+y}$. After more factorization we will get that $xy+zx+zy=0$. So, now it is easy to see $x^3+y^3-(x+y+z)(x^2+y^2)=2008$, so $xy^2+x^2y+x^2z+y^2z=2008$ and again after factorization we reach that $xyz=1004$