Slovenian Mathematical Olympiad, 2008

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

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

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

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