Example 4 Evaluate:
(1) $\cos 40^{\circ}+\cos 80^{\circ}+\cos 160^{\circ}$;
(2) $\cos 40^{\circ} \cdot \cos 80^{\circ}+\cos 80^{\circ} \cdot \cos 160^{\circ}+$ $\cos 160^{\circ} \cdot \cos 40^{\circ}$;
(3) $\cos 40^{\circ} \cdot \cos 80^{\circ} \cdot \cos 160^{\circ}$.
【Analysis】If we look at a single equation, it is quite difficult to find the value, and it requires a high level of proficiency in the application of trigonometric formulas. However, when considering the three equations together, their structure perfectly meets the requirements of Vieta's formulas, allowing us to construct a cubic equation to solve the problem.
From the special angle $120^{\circ}$ and the triple angle formula, we get
$$
\begin{array}{l}
\cos 120^{\circ}=\cos \left(3 \times 40^{\circ}\right) \\
=4 \cos ^{3} 40^{\circ}-3 \cos 40^{\circ}=-\frac{1}{2}, \\
\cos 240^{\circ}=\cos \left(3 \times 80^{\circ}\right) \\
=4 \cos ^{3} 80^{\circ}-3 \cos 80^{\circ}=-\frac{1}{2}, \\
\cos 480^{\circ}=\cos \left(3 \times 160^{\circ}\right) \\
=4 \cos ^{3} 160^{\circ}-3 \cos 160^{\circ}=-\frac{1}{2} .
\end{array}
$$

Thus, $\cos 40^{\circ} 、 \cos 80^{\circ} 、 \cos 160^{\circ}$ are the three roots of the equation $8 x^{3}-6 x+1=0$.
By Vieta's formulas, we get
$$
\begin{array}{l}
\cos 40^{\circ}+\cos 80^{\circ}+\cos 160^{\circ}=0 \\
\cos 40^{\circ} \cdot \cos 80^{\circ}+\cos 80^{\circ} \cdot \cos 160^{\circ}+ \\
\cos 160^{\circ} \cdot \cos 40^{\circ}=-\frac{3}{4} \\
\cos 40^{\circ} \cdot \cos 80^{\circ} \cdot \cos 160^{\circ}=-\frac{1}{8} .
\end{array}
$$