Example 3 In $\triangle A B C$, $A^{\prime} 、 B^{\prime} 、 C^{\prime}$ are on $B C 、 C A 、 A B$ respectively. Given that $A A^{\prime} 、 B B^{\prime} 、 C C^{\prime}$ concur at $O$, and $\frac{A O}{O A^{\prime}}+\frac{B O}{O B^{\prime}}+\frac{C O}{O C^{\prime}}=92$. Find the value of $\frac{A O}{O A^{\prime}} \cdot \frac{B O}{O B^{\prime}} \cdot \frac{C O}{O C^{\prime}}$.
(10th American Invitational Mathematics Examination)
Let $\angle A O B^{\prime}=\theta_{1}, \angle A O C^{\prime}=\theta_{2}$.
As shown in Figure 3, from $S_{\triangle A O B}+$
$S_{\triangle C O B^{\prime}}=S_{\triangle M O C}$, we get
$$
\begin{array}{l}
\frac{1}{2} \rho_{1} \rho_{2} \sin \theta_{1}+ \\
\frac{1}{2} \rho_{1} \rho_{6} \sin \left[180^{\circ}-\right. \\
\left.\left(\theta_{1}+\theta_{2}\right)\right] \\
=\frac{1}{2} \rho_{2} \rho_{6} \sin \left(180^{\circ}-\theta_{2}\right) .
\end{array}
$$

Then $\rho_{1} \rho_{2} \sin \theta_{1}-\rho_{2} \rho_{6} \sin \theta_{2}+$
$$
\rho_{1} \rho_{6} \sin \left(\theta_{1}+\theta_{2}\right)=0 \text {. }
$$

Similarly, we get
$$
\begin{array}{l}
-\rho_{2} \rho_{4} \sin \theta_{1}+\rho_{2} \rho_{3} \sin \theta_{2}+\rho_{3} \rho_{4} \sin \left(\theta_{1}+\theta_{2}\right)=0 . \\
\rho_{4} \rho_{5} \sin \theta_{1}+\rho_{5} \rho_{6} \sin \theta_{2}-\rho_{4} \rho_{6} \sin \left(\theta_{1}+\theta_{2}\right)=0 .
\end{array}
$$

Considering equations (1), (2), and (3) as a homogeneous linear system of equations in $\sin \theta_{1}, \sin \theta_{2}$, and $\sin \left(\theta_{1}+\theta_{2}\right)$, and knowing that there must be a non-zero solution, we have
$$
\left|\begin{array}{ccc}
\rho_{1} \rho_{2} & -\rho_{2} \rho_{6} & \rho_{1} \rho_{6} \\
-\rho_{2} \rho_{4} & \rho_{2} \rho_{3} & \rho_{3} \rho_{4} \\
\rho_{4} \rho_{5} & \rho_{5} \rho_{6} & -\rho_{4} \rho_{6}
\end{array}\right|=0 .
$$

Expanding and dividing both sides by $\rho_{2} \rho_{4} \rho_{6}$, we get
$$
\rho_{2} \rho_{4} \rho_{6}=\rho_{1} \rho_{2} \rho_{3}+2 \rho_{1} \rho_{3} \rho_{5}+\rho_{3} \rho_{4} \rho_{5}+\rho_{1} \rho_{5} \rho_{6} \text {. }
$$

Let $\rho_{2}=k_{1} \rho_{5}, \rho_{4}=k_{2} \rho_{1}, \rho_{6}=k_{3} \rho_{3}$.
Substituting into equation (4) and dividing by $\rho_{1} \rho_{3} \rho_{5}$, we get
$$
k_{1} k_{2} k_{3}=k_{1}+k_{2}+k_{3}+2 \text {. }
$$

Given $k_{1}+k_{2}+k_{3}=92$, we have $k_{1} k_{2} k_{3}=94$. Therefore, $\frac{A O}{O A^{\prime}} \cdot \frac{B O}{O B^{\prime}} \cdot \frac{C O}{O C^{\prime}}=94$.