Which pairs of numbers $(c ; d)$ are such that the equation $x^{3}-8 x^{2}+c x+d=0$ has three, not necessarily distinct, positive integer roots?
Solution. Let the roots of the equation be $e, f$ and $g$ (positive integers). This means that

$$
x^{3}-8 x^{2}+c x+d=(x-e)(x-f)(x-g)
$$

Performing the multiplication:

$$
\begin{gathered}
(x-e)(x-f)(x-g)=x^{3}-e x^{2}-f x^{2}-g x^{2}+e f x+f g x+e g x-e f g= \\
=x^{3}-(e+f+g) x^{2}+(e f+f g+e g) x-e f g
\end{gathered}
$$

Thus,

$$
x^{3}-8 x^{2}+c x+d=x^{3}-(e+f+g) x^{2}+(e f+f g+e g) x-e f g
$$

From this, we get that

$$
\begin{aligned}
e+f+g & =8 \\
e f+f g+e g & =c \\
e f g & =-d
\end{aligned}
$$

All three equations are symmetric, so we can assume that $e \geq f \geq g$.

Due to equation (1) (considering that $e, f, g$ are positive integers and $e \geq f \geq g$), the values of $e, f, g$ and the calculated values of $c$ and $d$ can only be the following:

|  | $e$ | $f$ | $g$ | $c$ | $d$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 1. | 6 | 1 | 1 | 13 | -6 |
| 2. | 5 | 2 | 1 | 17 | -10 |
| 3. | 4 | 3 | 1 | 19 | -12 |
| 4. | 4 | 2 | 2 | 20 | -16 |
| 5. | 3 | 3 | 2 | 21 | -18 |

Therefore, the appropriate $(c ; d)$ pairs are: $(13 ;-6),(17 ;-10),(19 ;-12),(20 ;-16),(21 ;-18)$.