55rd Ukrainian National Mathematical Olympiad - Third Round (Second Tour)

From the first equation $-1\le a\le 1$ and $-1\le b\le 1$, therefore $0\le 1+a\le 2$ and $0\le 1+b\le 2$. Add both equations and get $$a^2(1+a)+b^2(1+b)=0.$$ As both items are non-negative, their sum equals zero if and only if every item equals $0$. So $a,b\in \{-1,0\}$. From the first equation it follows that one variable is not zero, from the second equation it follows that this variable equals $-1$. Thus, the equations are satisfied by such pairs of numbers $(a,b):(-1,0)$ and $(0,-1)$, hence $a^3+b^2=-1$ or $a^3+b^2=1$.