Example 3 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.
Analysis: It can be observed that 1 is a solution to the equation. The equation can be transformed into
$$
(x-1)\left(5 x^{2}-5 p x+66 p-1\right)=0 \text {. }
$$

The problem is then reduced to: Find all real numbers $p$ such that the equation
$$
5 x^{2}-5 p x+66 p-1=0
$$

has natural number solutions.
By Vieta's formulas, $p$ is the sum of the roots of the equation, i.e., $p$ is a natural number. Following Example 2, we get
$$
\begin{array}{l}
\Delta=(5 p-132)^{2}-17404 . \\
\text { Let }(5 p-132)^{2}-17404=n^{2}(n>0, n \text { is a natural number}). \\
\end{array}
$$
Rearranging and factoring, we get
$$
\begin{array}{l}
(5 p-132+n)(5 p-132-n) \\
=2^{2} \times 19 \times 229 . \\
\text { Also, }(5 p-132+n),(5 p-132-n) \text { have the same parity, }
\end{array}
$$

Therefore,
$$
\left\{\begin{array}{l}
5 p-132+n=2 \times 229 \\
5 p-132-n=2 \times 19
\end{array}\right.
$$

Solving, we get $p=76$.
Note: On the surface, $p$ in this problem is any real number, but by Vieta's formulas, it is actually a natural number, so the method used above can be applied to find it.