## Task A-1.3.

Let $a$ and $b$ be positive real numbers such that $a^{2}+b^{2}=8$ and $a^{6}+b^{6}=416$.

Determine $a b$.
## Solution.

According to the formula for the cube of a binomial, we get:

$$
\begin{aligned}
\left(a^{2}+b^{2}\right)^{3} & =a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6} & & 1 \text { point } \\
& =a^{6}+b^{6}+3 a^{2} b^{2}\left(a^{2}+b^{2}\right) . & & 2 \text { points }
\end{aligned}
$$

By substituting the data from the problem into the above equality, we get:

$$
8^{3}=416+3 a^{3} b^{3} \cdot 8
$$

Rearranging gives $a^{2} b^{2}=4$.

Since $a$ and $b$ are positive, it follows that $a b>0$, so $a b=2$.

Note: The conditions of the problem uniquely determine $a$ and $b$. These are the numbers $\sqrt{3}-1$ and $\sqrt{3}+1$. If we wish, the numbers $a$ and $b$ can be determined as follows.

Once we know $a b=2$, we first calculate $a+b$ from $(a+b)^{2}=a^{2}+2 a b+b^{2}=12$. Then we get the system $a b=2, a+b=2 \sqrt{3}$ which leads to the quadratic equation $a^{2}-2 \sqrt{3} a+2=0$.

First-year students can solve the quadratic equation by the method of completing the square. However, in this problem, students are not required or expected to determine the numbers $a$ and $b$.