3.135. $-\cos 5 \alpha \cos 4 \alpha-\cos 4 \alpha \cos 3 \alpha+2 \cos ^{2} 2 \alpha \cos \alpha$.
Solution.

$-\cos 5 \alpha \cos 4 \alpha-\cos 4 \alpha \cos 3 \alpha+2 \cos ^{2} 2 \alpha \cos \alpha=$

$=-\cos 4 \alpha(\cos 5 \alpha+\cos 3 \alpha)+2 \cos ^{2} 2 \alpha \cos \alpha=$

$=-\cos 4 \alpha \cdot 2 \cos 4 \alpha \cos \alpha+2 \cos ^{2} 2 \alpha \cos \alpha=$

$=-2 \cos ^{2} 4 \alpha \cos \alpha+2 \cos ^{2} 2 \alpha \cos \alpha=-2 \cos \alpha\left(\cos ^{2} 4 \alpha-\cos ^{2} 2 \alpha\right)=$

$=-2 \cos \alpha(\cos 4 \alpha-\cos 2 \alpha)(\cos 4 \alpha+\cos 2 \alpha)=$

$=-2 \cos \alpha(-2 \sin 3 \alpha \sin \alpha) \cdot 2 \cos 3 \alpha \cos \alpha=$

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

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