Slovenija 2008

First solution Since $a^9+b^9=(a^3+b^3)(a^6-a^3{b}^3+b^6)$, we have $a^6-a^3{b}^3+b^6=-\frac{299}{13}=-23$.

Subtracting this equality from $a^6+2a^3{b}^3+b^6=(a^3+b^3{)}^2=13^2=169$ we get $3a^3{b}^3=192$, which implies $ab=4$ since $ab$ is a real number.

Second solution We have $13^3=(a^3+b^3{)}^3=a^9+3a^6{b}^3+3a^3{b}^6+b^9=-299+3a^6{b}^3+3a^3{b}^6$, which implies $832=a^6{b}^3+a^3{b}^6$.

Factoring the expression on the right-hand side we find that $832=a^3{b}^3(a^3+b^3)=a^3{b}^3\cdot 13$, or $a^3{b}^3=\frac{832}{13}=64$, so $ab=4$.