40th Hellenic Mathematical Olympiad

1. Since $c=-a-b$, we get: $$\begin{array}{rl}0& =a{b}^3+b(-a-b{)}^3+(-a-b)a^3\\ & =-(a{b}^3+b(a+b{)}^3+(a+b)a^3)\\ & =-(a^4+2a^{3}b+3a^2{b}^2+2a{b}^3+b^4)\\ & =-(a^2(a+b{)}^2+b^2(a+b{)}^2+a^2{b}^2)\\ & =-a^2{c}^2-b^2{c}^2-a^2{b}^2.\end{array}$$ Therefore each term of the last sum must be zero: $ab=bc=ca=0$. Hence two of the numbers must be zero and from the equality $a+b+c=0$, we conclude that $a=b=c=0$.