Example 6 Given three real numbers $x_{1}, x_{2}, x_{3}$, any one of these numbers plus five times the product of the other two always equals 6. The number of such triples $\left(x_{1}, x_{2}, x_{3}\right)$ is $\qquad$.
$(1995$, Dongfang Airlines Cup - Shanghai Junior High School Mathematics Competition)
Solving the system of equations given by the problem, we have:
$$
\left\{\begin{array}{l}
x_{1}+5 x_{2} x_{3}=6, \\
x_{2}+5 x_{3} x_{1}=6, \\
x_{3}+5 x_{1} x_{2}=6 .
\end{array}\right.
$$
(1) - (2) gives
$$
\left(x_{1}-x_{2}\right)\left(1-5 x_{3}\right)=0 \text {. }
$$
(2) - (3) gives
$$
\left(x_{2}-x_{3}\right)\left(1-5 x_{1}\right)=0 \text {. }
$$

From (1), (4), and (5), the original system of equations can be transformed into:
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ x _ { 1 } = x _ { 2 } , } \\
{ x _ { 2 } = x _ { 3 } , } \\
{ x _ { 3 } + 5 x _ { 1 } x _ { 2 } = 6 ; }
\end{array} \quad \left\{\begin{array}{l}
x_{1}=x_{2}, \\
5 x_{1}=1, \\
x_{3}+5 x_{1} x_{2}=6 ;
\end{array}\right.\right. \\
\left\{\begin{array} { l } 
{ 5 x _ { 3 } = 1 , } \\
{ x _ { 2 } = x _ { 3 } , } \\
{ x _ { 3 } + 5 x _ { 1 } x _ { 2 } = 6 ; }
\end{array} \quad \left\{\begin{array}{l}
5 x_{3}=1, \\
5 x_{1}=1, \\
x_{3}+5 x_{1} x_{2}=6 .
\end{array}\right.\right.
\end{array}
$$

Solving these four systems of equations, we get 5 sets of solutions for $\left(x_{1}, x_{2}, x_{3}\right)$: $\square$
$$
\begin{array}{l}
(1,1,1),\left(-\frac{6}{5},-\frac{6}{5},-\frac{6}{5}\right), \\
\left(\frac{1}{5}, \frac{1}{5}, \frac{29}{5}\right),\left(\frac{1}{5}, \frac{29}{5}, \frac{1}{5}\right),\left(\frac{29}{5}, \frac{1}{5}, \frac{1}{5}\right) .
\end{array}
$$