II. (This question is worth 25 points) 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.
From observation, it is easy to see that $x=1$ is a root of the original cubic equation. By synthetic division, the original cubic equation can be reduced to a quadratic equation:
$$
5 x^{2}-5 p x+66 p-1=0 \text {. }
$$

The three roots of the original cubic equation are natural numbers, and the two roots of the quadratic equation (*) are also natural numbers. Let $\alpha, \nu(2: \div \leqslant v)$ be the two roots of equation (*), 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 (1) into (2) gives $5 u v=66(u+v)-1$. It can be known that $u, v$ cannot be divisible by $2,3,11$.
From (3), we get $v=\frac{66 u-1}{5 u-66}$.
Since $u, v$ are natural numbers, from (4) we know $u>\frac{66}{5}$, i.e., $u \geqslant 14$.
Also, since $2 \nmid u$ and $3 \nmid u$, we know $u \geqslant 17$.
From $v \geqslant u$ and (4), we get $\frac{66 u-1}{5 u-66} \geqslant u$, i.e., $5 u^{2}-132 u+1 \leqslant 0$.
Thus, $u \leqslant \frac{66+\sqrt{66^{2}-5}}{5}<\frac{132}{5}$.
$\therefore 17 \leqslant u \leqslant 26$.
Furthermore, since $2 \nmid u$ and $3 \nmid u$, $u$ can only be $17,19,23,25$.
When $u=17$, from (4) we get $v=\frac{1121}{19}=59$.
When $u=19$, from (4) we get $v$ is not a natural number, so it should be discarded.
When $u=23$, from (4) we get $v$ is not a natural number, so it should be discarded.
When $u=25$, from (4) we get $v$ is not a natural number, so it should be discarded.
Therefore, only when $p=u+v=76$, the two roots of equation (*) are natural numbers, and the three roots of the original equation are natural numbers.