Example 8. In $\triangle A B C$, it is known that (1) $\operatorname{tg} A + \operatorname{tg} B + \operatorname{tg} C = -\frac{1}{6}$,
(2) $\operatorname{tg}^{3} A + \operatorname{tg}^{3} B + \operatorname{tg}^{3} C = -\frac{181}{216}$. Find the sizes of $A, B, C$.
$$
\begin{array}{c}
\text { Sol } \because A, B, C \text { are the three interior angles of } \triangle A B C, \\
\therefore \operatorname{tg} A+\operatorname{tg} B+\operatorname{tg} C=\operatorname{tg} A \operatorname{tg} B \operatorname{tg} C . \\
\because \operatorname{tg}^{3} A+\operatorname{tg}^{3} B+\operatorname{tg}{ }^{3} C-3 \operatorname{tg} A \operatorname{tg} B \operatorname{tg} C \\
=(\operatorname{tg} A+\operatorname{tg} B+\operatorname{tg} C)\left(\operatorname{tg}^{2} A+\operatorname{tg}{ }^{2} B\right. \\
\left.+\operatorname{tg}^{2} C-\operatorname{tg} A \operatorname{tg} B-\operatorname{tg} B \operatorname{tg} C-\operatorname{tg} C \operatorname{tg} A\right) \\
=(\operatorname{tg} A+\operatorname{tg} B+\operatorname{tg} C) C(\operatorname{tg} A+\operatorname{tg} B+\operatorname{tg} C)^{2} \\
-3(\operatorname{tg} A \operatorname{tg} B+\operatorname{tg} B \operatorname{tg} C+\operatorname{tg} C \operatorname{tg} A)], \\
\text { Substituting the known values, we simplify to get, } \operatorname{tg} A \operatorname{tg} B+\operatorname{tg} B \operatorname{tg} C \\
+\operatorname{tg} C \operatorname{tg} A=-\frac{2}{3} . \\
\therefore\left\{\begin{array}{c}
\operatorname{tg} A+\operatorname{tg} B+\operatorname{tg} C=-\frac{1}{6}, \\
\operatorname{tg} A \operatorname{tg} B+\operatorname{tg} B \operatorname{tg} C+\operatorname{tg} C \operatorname{tg} A \\
=-\frac{2}{3}, \\
\operatorname{tg} A \operatorname{tg} B \operatorname{tg} C=-\frac{1}{6} .
\end{array}\right.
\end{array}
$$

By the inverse of Vieta's Theorem: $\operatorname{tg} A, \operatorname{tg} B, \operatorname{tg} C$ are the three roots of the equation $x^{3}+\frac{1}{6} x^{2}-\frac{2}{3} x+\frac{1}{6}=0$ which is $6 x^{3}+x^{2}$ $-4 x+1=0$ or, $(x+1)(2 x-1)(3 x-1)$ $=0$.
$$
\because \quad x_{1}=-1, \quad x_{2}=\frac{1}{2}, \quad x_{3}=\frac{1}{3},
$$
$\therefore$ The sizes of the three interior angles are $\frac{3 \pi}{4}, \operatorname{arctg}_{2}^{-1}$,
$$
\operatorname{arctg} \frac{1}{3}
$$
$$