In three cells of a $3 \times 3$ table, numbers are placed (see the figure). It is required to fill in the remaining cells with numbers so that the sums of the numbers in all rows, columns, and main diagonals are equal. Prove that this can be done in a unique way, and fill in the table.

![](https://cdn.mathpix.com/cropped/2024_05_06_c07b2c4be0d3dcbcc4e5g-31.jpg?height=166&width=166&top_left_y=563&top_left_x=952)
The first method. Let the number in the empty cell of the top row be $x$, then the sum of the numbers in this row is 6 $+x$. For the sum of the numbers in the first column to be the same, the number in the empty cell of this column should be $x+2$. Considering the diagonal from the bottom left cell to the top right and reasoning similarly, we get that the number in the central cell should be $x-2$. Then to the right of it is the number $6-x$, and below it is the number $8-x$ (see the left diagram).

For the sums of the numbers in the third row and the third column to be equal to $6+x$, the number in the bottom right cell should be $2x-5$. On the other hand, the number 7 should be there for the sum of the numbers in the second diagonal to be the same. Therefore, $2x-5=7$, from which $x=6$. Thus, the table is uniquely restored (see the right diagram), as required.

| 1 | $x$ | 5 |
| :---: | :---: | :---: |
| $x+2$ | $x-2$ | $6-x$ |
| 3 | $8-x$ | $2 x-5$ |


| 1 | 6 | 5 |
| :--- | :--- | :--- |
| 8 | 4 | 0 |
| 3 | 2 | 7 |

The second method. Let the specified sums be $S$. Adding the sums of the numbers in the second column and along the diagonals and subtracting from this the sums of the numbers in the first and third rows, we get $S$. On the other hand, this number is equal to three times the number in the central cell.

On one of the diagonals are the numbers 3 and 5, so the number in the central cell is 4, and $S=12$. Now all the numbers in the table are easily restored.