In triangle $ABC$, $A'$, $B'$, and $C'$ are on the sides $BC$, $AC$, and $AB$, respectively. Given that $AA'$, $BB'$, and $CC'$ are concurrent at the point $O$, and that \[\frac{AO}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'}=92,\] find \[\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}.\]
1. Let \( \frac{AO}{OA'} = k_1 \), \( \frac{BO}{OB'} = k_2 \), and \( \frac{CO}{OC'} = k_3 \). We are given that:
   \[
   k_1 + k_2 + k_3 = 92
   \]
2. By the properties of cevians in a triangle, we know that:
   \[
   \frac{1}{k_1 + 1} + \frac{1}{k_2 + 1} + \frac{1}{k_3 + 1} = 1
   \]
3. To find \( k_1 k_2 k_3 \), we start by expressing the given equation in terms of \( k_1, k_2, \) and \( k_3 \):
   \[
   \frac{1}{k_1 + 1} + \frac{1}{k_2 + 1} + \frac{1}{k_3 + 1} = 1
   \]
4. Let \( x = k_1 + 1 \), \( y = k_2 + 1 \), and \( z = k_3 + 1 \). Then the equation becomes:
   \[
   \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1
   \]
5. We know that:
   \[
   x - 1 + y - 1 + z - 1 = 92 \implies x + y + z - 3 = 92 \implies x + y + z = 95
   \]
6. Using the identity for the sum of reciprocals, we multiply both sides by \( xyz \):
   \[
   yz + xz + xy = xyz
   \]
7. Substituting \( x = k_1 + 1 \), \( y = k_2 + 1 \), and \( z = k_3 + 1 \), we get:
   \[
   (k_1 + 1)(k_2 + 1) + (k_2 + 1)(k_3 + 1) + (k_3 + 1)(k_1 + 1) = (k_1 + 1)(k_2 + 1)(k_3 + 1)
   \]
8. Expanding and simplifying:
   \[
   (k_1 k_2 + k_1 + k_2 + 1) + (k_2 k_3 + k_2 + k_3 + 1) + (k_3 k_1 + k_3 + k_1 + 1) = k_1 k_2 k_3 + k_1 k_2 + k_2 k_3 + k_3 k_1 + k_1 + k_2 + k_3 + 1
   \]
9. Collecting like terms:
   \[
   k_1 k_2 + k_2 k_3 + k_3 k_1 + k_1 + k_2 + k_3 + 3 = k_1 k_2 k_3 + k_1 k_2 + k_2 k_3 + k_3 k_1 + k_1 + k_2 + k_3 + 1
   \]
10. Simplifying further:
    \[
    3 = k_1 k_2 k_3 + 1 \implies k_1 k_2 k_3 = 2
    \]

The final answer is \( \boxed{2} \).