Example 3 In $\triangle A B C$, it is known that $\tan A+\tan B+\tan C=\frac{1}{6}, \tan ^{3} A+\tan ^{3} B+\tan ^{3} C=-\frac{181}{216}$, find the sizes of $A, B, C$.
Solve: In $\triangle ABC$, we have $\tan A + \tan B + \tan C = \tan A \tan B \tan C$.
And $\tan^3 A + \tan^3 B + \tan^3 C - 3 \tan A \tan B \tan C = (\tan A + \tan B + \tan C)\left(\tan^2 A + \tan^2 B + \tan^2 C - \tan A \tan B - \tan B \tan C - \tan C \tan A\right) = (\tan A + \tan B + \tan C)\left[(\tan A + \tan B + \tan C)^2 - 3(\tan A \tan B + \tan B \tan C + \tan C \tan A)\right]$.
Substituting the known conditions, we get $\tan A \tan B + \tan B \tan C + \tan C \tan A = -\frac{2}{3}$.
Thus, $\left\{\begin{array}{l}\tan A + \tan B + \tan C = -\frac{1}{6} \\ \tan A \tan B + \tan B \tan C + \tan C \tan A = -\frac{2}{3} \\ \tan A \tan B \tan C = -\frac{1}{6}\end{array}\right.$
Therefore, $\tan A, \tan B, \tan C$ are the roots of the equation $x^3 + \frac{1}{6} x^2 - \frac{2}{3} x + \frac{1}{6} = 0$,
and $x^3 + \frac{1}{6} x^2 - \frac{2}{3} x + \frac{1}{6} = \frac{1}{6}(x + 1)(2x - 1)(3x - 1) = 0$,
so $x_1 = -1, x_2 = \frac{1}{2}, x_3 = \frac{1}{3}$. That is, $\{\tan A, \tan B, \tan C\} = \left\{-1, \frac{1}{2}, \frac{1}{3}\right\}$.
Therefore, the sizes of the three interior angles of $\triangle ABC$ are $\frac{3}{4} \pi, \arctan \frac{1}{2}, \arctan \frac{1}{3}$.