In a certain sequence of 80 numbers, any term, except for the two endpoints, is equal to the product of its neighboring terms. The product of the first 40 terms of the sequence is 8 and the product of all the terms is also 8. Determine the terms of the sequence.
Let $a_{1}, a_{2}, \ldots, a_{80}$ be the numbers of this sequence. For each $i \geq 1$, we have

$$
\left\{\begin{array}{l}
a_{i+1}=a_{i} \cdot a_{i+2} \\
a_{i+2}=a_{i+1} \cdot a_{i+3}
\end{array}\right.
$$

Consequently, $a_{i+1}=a_{i} \cdot a_{i+1} \cdot a_{i+3}$ and, since $a_{i+1} \neq 0$, as the product of the terms of the sequence is $8 \neq 0$, it follows that $a_{i} \cdot a_{i+3}=1$.

Any two numbers in the sequence, whose indices differ by 3, have a product equal to 1. Therefore, the product of any six consecutive numbers in this sequence is always equal to 1. Given that the product of the first 40 terms of the sequence is 8, we conclude that the product of the first four terms is also 8, because the remaining 36 terms form six groups of six consecutive terms of the sequence, and in each of these groups, the product is equal to 1. That is, $a_{1} a_{2} a_{3} a_{4}=8$. Since $a_{i} \cdot a_{i+3}=1$, it follows that $a_{1} a_{4}=1$ and, therefore, $a_{2} a_{3}=8$.

We also have the hypothesis that the product of the 80 terms of the sequence is 8, from which we can conclude that $a_{1} a_{2}=8$, since the last 78 terms can be grouped into 13 groups of six consecutive terms, each with a product equal to 1, as we have already seen.

Thus, from $a_{2} a_{3}=8$, $a_{1} a_{2}=8$, and $a_{2}=a_{1} a_{3}$, it follows that $a_{1} a_{2}^{2} a_{3}=64$ and $a_{2}^{3}=64$. Therefore,

$$
a_{1}=2, a_{2}=4 \text{ and } a_{3}=2
$$

Observe, also, that the entire sequence is now determined. Indeed, we have

$$
2,4,2, \frac{1}{2}, \frac{1}{4}, \frac{1}{2}, 2,4,2, \frac{1}{2}, \frac{1}{4}, \frac{1}{2}, \ldots
$$

where the first six terms repeat, always in the same order.