Sure, here is the translated text:

```
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 natural number root of the original cubic equation. By synthetic division, the original cubic equation can be reduced to a quadratic equation:
$$
5 x^{2}-5 p x+66 p-1=0
$$

The three roots of the original cubic equation are all natural numbers $\Leftrightarrow$ the two roots of the quadratic equation $\circledast$ are both natural numbers.
Let $u, v(u \leqslant v)$ 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 is known that $u$ and $v$ cannot be divisible by $2$, $3$, or $11$.
From (3) we get
$$
v=\frac{66 u-1}{5 u-66},
$$

Since $u$ and $v$ are both natural numbers, from (4) we know
$$
u>\frac{66}{5} \text { i.e. } u \geqslant 14 \text {. }
$$

Since $2 \nmid u$ and $3 \nmid u$, $\therefore \quad u \geqslant 17$,
From $v \geqslant u$ and (4) we get
$$
\frac{66 u-1}{5 u-66} \geqslant u,
$$

which means
$$
5 u^{2}-132 u+1 \leqslant 0 \text {. }
$$

Thus,
$$
u \leqslant \frac{66+\sqrt{66^{2}-5}}{5}<\frac{132}{5} \text {, }
$$
$\therefore$
$$
17 \leqslant u \leqslant 26 .
$$

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 $v=\frac{1253}{29}$ which is not a natural number, so it should be discarded.
When $u=23$, from (4) we get $v=\frac{1517}{49}$ which is not a natural number, so it should be discarded.
When $u=25$, from (4) we get $v=\frac{1649}{59}$ which is not a natural number, so it should be discarded.
$\therefore$ Only when $p=u+v=76$, the two roots of equation $*$ are both natural numbers, and the three roots of the original equation are all natural numbers.