Find all pairs $(x, y)$ of real numbers such that $|x|+|y|=1340$ and $x^{3}+y^{3}+2010 x y=670^{3}$.
Answer: $(-670 ;-670),(1005 ;-335),(-335 ; 1005)$.

To prove this, let $z=-670$. We have

$$
0=x^{3}+y^{3}+z^{3}-3 x y z=\frac{1}{2}(x+y+z)\left((x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right)
$$

Thus either $x+y+z=0$, or $x=y=z$. In the latter case we get $x=y=-670$, which satisfies both the equations. In the former case we get $x+y=670$. Then at least one of $x, y$ is positive, but not both, as from the second equation we would get $x+y=1340$. If $x>0 \geq y$, we get $x-y=1340$, which together with $x+y=670$ yields $x=1005, y=-335$. If $y>0 \geq x$ we get similarly $x=-335, y=1005$.