22. Let the polynomial $P(x)=x^{3}+a x^{2}+b x+c$ have roots $\cos \frac{2 \pi}{7}, \cos \frac{4 \pi}{7}, \cos \frac{6 \pi}{7}$. Then the value of $a b c$ is
(A) $-\frac{3}{49}$
(B) $-\frac{1}{28}$
(C) $\frac{\sqrt[3]{7}}{64}$
(D) $\frac{1}{32}$
(E) $\frac{1}{28}$
22. D.

By Vieta's formulas, we have
$$
\begin{array}{l}
\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}=-a, \\
\cos \frac{2 \pi}{7} \cdot \cos \frac{4 \pi}{7}+\cos \frac{4 \pi}{7} \cdot \cos \frac{6 \pi}{7}+\cos \frac{6 \pi}{7} \cdot \cos \frac{2 \pi}{7}=b, \\
-\cos \frac{2 \pi}{7} \cdot \cos \frac{4 \pi}{7} \cdot \cos \frac{6 \pi}{7}=c .
\end{array}
$$

Notice that,
$$
\begin{array}{l}
1+\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}+\cos \frac{8 \pi}{7}+ \\
\cos \frac{10 \pi}{7}+\cos \frac{12 \pi}{7}=0 \\
\Rightarrow 1+2\left(\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}\right)=0 \\
\Rightarrow a=\frac{1}{2} . \\
\text { By } \frac{1}{2}\left(\cos \frac{6 \pi}{7}+\cos \frac{2 \pi}{7}+\cos \frac{10 \pi}{7}+\cos \frac{2 \pi}{7}+\right. \\
\left.\cos \frac{8 \pi}{7}+\cos \frac{4 \pi}{7}\right)=b \\
\Rightarrow-\frac{1}{2}=\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}=b \\
\Rightarrow b=-\frac{1}{2} .
\end{array}
$$

By $\cos \frac{2 \pi}{7} \cdot \cos \frac{4 \pi}{7} \cdot \cos \frac{6 \pi}{7}$
$$
\begin{array}{l}
=\frac{-\cos \frac{\pi}{7} \cdot \cos \frac{2 \pi}{7} \cdot \cos \frac{4 \pi}{7} \cdot \sin \frac{\pi}{7} \cdot 2^{3}}{8 \sin \frac{\pi}{7}} \\
=-\frac{\sin \frac{8 \pi}{7}}{8 \sin \frac{\pi}{7}}=\frac{\sin \frac{\pi}{7}}{8 \sin \frac{\pi}{7}}=\frac{1}{8} \\
\Rightarrow c=-\frac{1}{8} .
\end{array}
$$

Therefore, $a b c=\frac{1}{2} \times\left(-\frac{1}{2}\right)\left(-\frac{1}{8}\right)=\frac{1}{32}$.