49th Mathematical Olympiad in Ukraine

From the first two equations of our system we obtain: $x^3-y^3=2y^3-2z^3+y-z=(y-z)(2y^2+2yz+2z^2+1)$, analogously, we can easily get equalities: $y^3-z^3=(z-x)(2z^2+2zx+2x^2+1)$, $z^3-x^3=(x-y)(2x^2+2xy+2y^2+1)$. It is easy to see that $(2y^2+2yz+2z^2+1)=y^2+z^2+(y+z{)}^2+1>0$. If we suppose that $y>z$, then from the first equality we have that $x>y$, then applying third equality we conclude: $z>x>y>z$ - contradiction. The case when $y<z$ is completely analogous. Hence we obtain $x=y=z$. and solving the equation we get the answer $(1,1,1)$.