322. Express $\operatorname{tg} 6 \alpha$ in terms of $\operatorname{tg} \alpha$.
322. According to the formulas of problem 321 b), we have:

$$
\operatorname{tg} 6 \alpha=\frac{\sin 6 \alpha}{\cos 6 \alpha}=\frac{6 \cos ^{5} \alpha \sin \alpha-20 \cos ^{3} \alpha \sin ^{3} \alpha+6 \cos \alpha \sin ^{5} \alpha}{\cos ^{6} \alpha-15 \cos ^{4} \alpha \sin ^{2} \alpha+15 \cos ^{2} \alpha \sin ^{4} \alpha-\sin ^{6} \alpha}
$$

Dividing the numerator and the denominator of the last fraction by $\cos ^{6} \alpha$, we obtain the following formula:

$$
\operatorname{tg} 6 \alpha=\frac{6 \operatorname{tg} \alpha-20 \operatorname{tg}^{3} \alpha+6 \operatorname{tg}^{5} \alpha}{1-15 \operatorname{tg}^{2} \alpha+15 \operatorname{tg}^{4} \alpha-\operatorname{tg}^{6} \alpha}
$$