$3.410 \sin ^{3} 2 \alpha \cdot \cos 6 \alpha+\cos ^{3} 2 \alpha \cdot \sin 6 \alpha$.
## Solution.

$(\sin 6 \alpha \cdot \cos 2 \alpha) \cdot \cos ^{2} 2 \alpha+(\cos 6 \alpha \cdot \sin 2 \alpha) \cdot \sin ^{2} 2 \alpha=\sin 6 \alpha \cdot \cos 2 \alpha \cdot \frac{1+\cos 4 \alpha}{2}+$ $+\cos 6 \alpha \cdot \sin 2 \alpha \cdot \frac{1-\cos 4 \alpha}{2}=\frac{1}{2}[(\sin 6 \alpha \cdot \cos 2 \alpha+\cos 6 \alpha \cdot \sin 2 \alpha)+\cos 4 \alpha \times$ $\times(\sin 6 \alpha \cdot \cos 2 \alpha-\cos 6 \alpha \cdot \sin 2 \alpha)]=\frac{1}{2}(\sin 8 \alpha+\cos 4 \alpha \cdot \sin 4 \alpha)=$ $=\frac{1}{2}\left(\sin 8 \alpha+\frac{\sin 8 \alpha}{2}\right)=\frac{3}{4} \sin 8 \alpha$.