It is easy to show that for every whole number $a$ holds $3|a^3 - a$, and that the square of a number modulo $3$ can be $0$ or $1$.

Now the given equation is equivalent to $x^3 - x + 2(y^3 - y) - 3(x - y) + z^2 = 2012$. From the above concluded and from the fact that $2012 \equiv 2 \pmod{3}$ we get that $z^2 \equiv 2 \pmod{3}$, which is impossible. Hence the given equation has no solution in the set of whole numbers.