Answer: There are three triples: $(2,3,5),(3,5,7),(2,3,7)$.

Solution. Let's call a face of the cube beautiful if one of the numbers on it is equal to the sum of the other three. Let's call a number large if on a beautiful face it is equal to the sum of the other three numbers. Clearly, the large numbers can only be $6,7,8$, and

- in the case of the large number 6, it is on the same face with the numbers 1, 2, 3;
- in the case of the large number 7, it is on the same face with the numbers 1, 2, 4;
- in the case of the large number 8, it is on the same face with the numbers $1,2,5$ or $1,3,4$.

Clearly, two large numbers cannot be on the same beautiful face at the same time. Also, it is clear that for each large number, the number 1 is on the same face with it, so all three beautiful faces have a common vertex, which is 1.

Two vertices are called adjacent if they are connected by an edge of the cube. Let's consider two cases.

- The numbers 1 and 8 are not in adjacent vertices. Clearly, the numbers 6 or 7 cannot be in the vertices adjacent to 1 (otherwise, two large numbers will be on the same beautiful face). Therefore, in all three beautiful faces, the large numbers are different (i.e., they are $6,7,8$) and are opposite to 1 (i.e., not adjacent to the vertex 1). The beautiful faces with the numbers 6 and 7 have two common vertices, which are definitely 1 and 2, then the beautiful faces with the numbers 6 and 8 have two common vertices, which are 1 and 3, and the beautiful faces with the numbers 7 and 8 have two common vertices, which are 1 and 4. Therefore, in the only vertex not belonging to the three beautiful faces, there is 5. Thus, in the vertices adjacent to 6, there are the numbers $2,3,5$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8af0c885427e3e323cf9g-31.jpg?height=301&width=282&top_left_y=697&top_left_x=628)

- The numbers 1 and 8 are in adjacent vertices. Then the number 8 is large in two beautiful faces at the same time. Then, on the edge connecting 1 and 8, in one face there are the numbers 3 and 4, and in the other - 2 and 5. In the remaining beautiful face, the large number is opposite to 1, and it is either 6 or 7.
- If it is 6, then in this face there are the numbers 2 and 3, and the numbers 4 and 5 lie in the opposite face of the cube, together with the remaining number 7. Thus, in the vertices adjacent to 6, there are the numbers $2,3,7$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8af0c885427e3e323cf9g-31.jpg?height=303&width=283&top_left_y=1356&top_left_x=670)

- If it is 7, then in this face there are the numbers 2 and 4, and the numbers 3 and 5 lie in the opposite face of the cube, together with the remaining number 6. Thus, in the vertices adjacent to 6, there are the numbers $3,5,7$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8af0c885427e3e323cf9g-32.jpg?height=301&width=291&top_left_y=92&top_left_x=662)
Problem 9.2. In the vertices of a cube, numbers $1,2, \ldots, 8$ are written in some order. It turned out that on three faces of the cube, the following condition is met: one of the numbers at the vertices is equal to the sum of the other three.

From the vertex with the number 6, three edges emanate. What three numbers can stand at their ends? The answers can be entered in any order. It is sufficient to provide one suitable example.