3.280. $\frac{\sin ^{2} 4 \alpha+4 \sin ^{4} 2 \alpha-4 \sin ^{2} 2 \alpha \cos ^{2} 2 \alpha}{4-\sin ^{2} 4 \alpha-4 \sin ^{2} 2 \alpha}$.
## Solution.

$$
\begin{aligned}
& \frac{\sin ^{2} 4 \alpha+4 \sin ^{4} 2 \alpha-4 \sin ^{2} 2 \alpha \cos ^{2} 2 \alpha}{4-\sin ^{2} 4 \alpha-4 \sin ^{2} 2 \alpha}= \\
& =\frac{(\sin 2(2 \alpha))^{2}+4 \sin ^{4} 2 \alpha-4 \sin ^{2} 2 \alpha \cos ^{2} 2 \alpha}{\left(4-4 \sin ^{2} 4 \alpha\right)-(\sin 2(2 \alpha))^{2}}= \\
& =\frac{(2 \sin 2 \alpha \cos 2 \alpha)^{2}+4 \sin ^{4} 2 \alpha-4 \sin ^{2} 2 \alpha \cos ^{2} 2 \alpha}{4\left(1-\sin ^{2} 2 \alpha\right)-(2 \sin 2 \alpha \cos 2 \alpha)^{2}}= \\
& =\frac{4 \sin ^{2} 2 \alpha \cos ^{2} 2 \alpha+4 \sin ^{4} 2 \alpha-4 \sin ^{2} 2 \alpha \cos ^{2} 2 \alpha}{4 \cos ^{2} 2 \alpha-4 \sin ^{2} 2 \alpha \cos ^{2} 2 \alpha}= \\
& =\frac{4 \sin ^{4} 2 \alpha}{4 \cos ^{2} 2 \alpha\left(1-\sin ^{2} 2 \alpha\right)}=\frac{\sin ^{4} 2 \alpha}{\cos ^{2} 2 \alpha \cos ^{2} 2 \alpha}=\frac{\sin ^{4} 2 \alpha}{\cos ^{4} 2 \alpha}=\tan^{4} 2 \alpha
\end{aligned}
$$

Answer: $\tan^{4} 2 \alpha$.