$3.4113 \sin \alpha \cdot \cos 3 \alpha+9 \sin \alpha \cdot \cos \alpha-\sin 3 \alpha \cdot \cos 3 \alpha-3 \sin 3 \alpha \cdot \cos \alpha$.
Solution. $3(\sin \alpha \cdot \cos 3 \alpha-\cos \alpha \cdot \sin 3 \alpha)+\frac{9}{2} \sin 2 \alpha-\frac{\sin 6 \alpha}{2}=$ $=-3 \sin 2 \alpha+\frac{9}{2} \sin 2 \alpha-\frac{3 \sin 2 \alpha-4 \sin ^{3} 2 \alpha}{2}=2 \sin ^{3} 2 \alpha$.

$3.4124\left(\sin ^{4} \alpha+\cos ^{4} \alpha\right)-4\left(\sin ^{6} \alpha+\cos ^{6} \alpha\right)-1$.

## Solution.

$$
\begin{aligned}
& 4\left(\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)^{2}-2 \sin ^{2} \alpha \cdot \cos ^{2} \alpha\right)-4\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)\left(\sin ^{4} \alpha-\sin ^{2} \alpha \times\right. \\
& \left.\times \cos ^{2} \alpha+\cos ^{4} \alpha\right)-1=4-2 \sin ^{2} 2 \alpha-4\left(\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)^{2}-3 \sin ^{2} \alpha \cdot \cos ^{2} \alpha\right)-1= \\
& =4-2 \sin ^{2} 2 \alpha-4+3 \sin ^{2} 2 \alpha-1=-\cos ^{2} 2 \alpha
\end{aligned}
$$

## Transform into a product (3.413 - 3.415):