3.136. $\sin 10 \alpha \sin 8 \alpha+\sin 8 \alpha \sin 6 \alpha-\sin 4 \alpha \sin 2 \alpha$.
Solution.

$\sin 10 \alpha \sin 8 \alpha+\sin 8 \alpha \sin 6 \alpha-\sin 4 \alpha \sin 2 \alpha .=$

$=\sin 8 \alpha(\sin 10 \alpha+\sin 6 \alpha)-\sin 4 \alpha \sin 2 \alpha=$

$=2 \sin 8 \alpha \sin 8 \alpha \cos 2 \alpha-\sin 4 \alpha \sin 2 \alpha=$
$=2 \sin ^{2} 8 \alpha \cos 2 \alpha-\sin 4 \alpha \sin 2 \alpha=$

$=2 \sin ^{2} 8 \alpha \cos 2 \alpha-2 \sin 2 \alpha \cos 2 \alpha \sin 2 \alpha=$

$=2 \sin ^{2} 8 \alpha \cos 2 \alpha-2 \sin ^{2} 2 \alpha \cos 2 \alpha=2 \cos 2 \alpha\left(\sin ^{2} 8 \alpha-\sin ^{2} 2 \alpha\right)=$

$=2 \cos 2 \alpha(\sin 8 \alpha-\sin 2 \alpha)(\sin 8 \alpha+\sin 2 \alpha)=$

$=2 \cos 2 \alpha \cdot 2 \cos 5 \alpha \sin 3 \alpha \cdot 2 \sin 5 \alpha \cos 3 \alpha=$

$=2 \cos 2 \alpha(2 \sin 3 \alpha \cos 3 \alpha)(2 \sin 5 \alpha \cos 5 \alpha)=2 \cos 2 \alpha \sin 6 \alpha \sin 10 \alpha$.

Answer: $2 \cos 2 \alpha \sin 6 \alpha \sin 10 \alpha$.