$5 \cdot 81$ Find all real numbers $P$ such that the cubic equation
$$
5 x^{3}-5(p+1) x^{2}+(71 p-1) x+1=66 p
$$

has three roots that are all natural numbers.
[Solution] By observation, $x=1$ is a positive integer root of the given cubic equation.
Using synthetic division to reduce the original cubic equation to a quadratic equation
$$
5 x^{2}-5 p x+66 p-1=0
$$

Obviously, the necessary and sufficient condition for the three roots of the original cubic equation to be natural numbers is that the two roots of the quadratic equation (1) are natural numbers.
Let the two natural number roots of equation (1) be $u$ and $v(u \leqslant v)$. Then by Vieta's formulas, we have
$$
\left\{\begin{array}{l}
u+v=p, \\
u v=\frac{1}{5}(66 p-1) .
\end{array}\right.
$$

Substituting (2) into (3) and eliminating $p$, we get
$$
5 u v=66(u+v)-1 .
$$

The above equation indicates that $u$ and $v$ are not multiples of $2, 3, 11$.
From (4), we get
$$
v=\frac{66 u-1}{5 u-66} .
$$

Since $u, v$ are natural numbers, it follows from the above equation that
$$
\begin{array}{l}
5 u-66>0, \\
u \geqslant 14 .
\end{array}
$$

Since $u$ is not a multiple of 2 and 3, we have
$$
u \geqslant 17 \text {. }
$$

From $u \leqslant v$, we get
$$
\begin{array}{l}
u \leqslant \frac{66 u-1}{5 u-66}, \\
5 u^{2}-132 u+1 \leqslant 0, \\
u \leqslant \frac{66+\sqrt{66^{2}-5}}{5}<\frac{132}{5},
\end{array}
$$

Therefore, $\quad 17 \leqslant u \leqslant 26$.

Furthermore, since $2 \times u$ and $3 \times u$, we have $u \in\{17,19,23,25\}$.
If $u=17$, then $v=\frac{1121}{19}=59$.
If $u=19,23$ or 25, then $v$ is not a natural number.
Therefore, only when $p=u+v=17+59=76$, the two roots of equation (1) are natural numbers, and thus the three roots of the original equation are natural numbers.