3. For numbers $a$ and $b$, it holds that $a^{3}+b^{3}=13$ and $a^{9}+b^{9}=-299$. What is $a b$, given that the number $a b$ is real?
II/3. 1. way Since $a^{9}+b^{9}=\left(a^{3}+b^{3}\right)\left(a^{6}-a^{3} b^{3}+b^{6}\right)$, we have $a^{6}-a^{3} b^{3}+b^{6}=-\frac{299}{13}=-23$. We also note that $a^{6}+2 a^{3} b^{3}+b^{6}=\left(a^{3}+b^{3}\right)^{2}=13^{2}=169$. Subtracting the equations, we get $3 a^{3} b^{3}=192$, from which it follows that $a b=4$, since $a b$ is a real number.
Factorization $a^{9}+b^{9}=\left(a^{3}+b^{3}\right)\left(a^{6}-a^{3} b^{3}+b^{6}\right)$ ..... 2 points
Conclusion $a^{6}-a^{3} b^{3}+b^{6}=-23$ ..... 1 point
Expression $\left(a^{3}+b^{3}\right)^{2}=a^{6}+2 a^{3} b^{3}+b^{6}$ ..... 2 points
Conclusion $3 a^{3} b^{3}=192$ (or equivalent) ..... 1 point
Answer $a b=4$ ..... 1 point

2. way We have $13^{3}=\left(a^{3}+b^{3}\right)^{3}=a^{9}+3 a^{6} b^{3}+3 a^{3} b^{6}+b^{9}=-299+3 a^{6} b^{3}+3 a^{3} b^{6}$, from which it follows that $832=a^{6} b^{3}+a^{3} b^{6}$. If we factor the expression on the right, we get $832=a^{3} b^{3}\left(a^{3}+b^{3}\right)=a^{3} b^{3} \cdot 13$, from which it follows that $a^{3} b^{3}=\frac{832}{13}=64$, so $a b=4$, since $a b$ is a real number.

Expression $\left(a^{3}+b^{3}\right)^{3}=a^{9}+3 a^{6} b^{3}+3 a^{3} b^{6}+b^{9}$

2 points

Calculation $832=a^{6} b^{3}+a^{3} b^{6}$ (or equivalent)

1 point

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