3.270. $\frac{\sin 2 \alpha+\cos 2 \alpha-\cos 6 \alpha-\sin 6 \alpha}{\sin 4 \alpha+2 \sin ^{2} 2 \alpha-1}$.
Solution.

$$
\begin{aligned}
& \frac{\sin 2 \alpha+\cos 2 \alpha-\cos 6 \alpha-\sin 6 \alpha}{\sin 4 \alpha+2 \sin ^{2} 2 \alpha-1}= \\
& =\frac{(\sin 2 \alpha-\sin 6 \alpha)+(\cos 2 \alpha-\cos 6 \alpha)}{\sin 4 \alpha-\left(1-2 \sin ^{2} 2 \alpha\right)}= \\
& =\left[\sin x-\sin y=2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}\right. \\
& \left.\cos x-\cos y=-2 \sin \frac{x+y}{2} \sin \frac{x-y}{2} ; 1-2 \sin ^{2} x=\cos 2 x\right]= \\
& =\frac{2 \cos 4 \alpha(-\sin 2 \alpha)+(-2 \sin 4 \alpha(-\sin 2 \alpha))}{\sin 4 \alpha-\cos 4 \alpha}= \\
& =\frac{-2 \cos 4 \alpha \sin 2 \alpha+2 \sin 4 \alpha \sin 2 \alpha}{\sin 4 \alpha-\cos 4 \alpha}=\frac{2 \sin 2 \alpha(\sin 4 \alpha-\cos 4 \alpha)}{\sin 4 \alpha-\cos 4 \alpha}=2 \sin 2 \alpha .
\end{aligned}
$$

Answer: $2 \sin 2 \alpha$.