3.255. $\cos ^{4} 2 \alpha-6 \cos ^{2} 2 \alpha \sin ^{2} 2 \alpha+\sin ^{4} 2 \alpha$.
Solution.

$$
\begin{aligned}
& \cos ^{4} 2 \alpha-6 \cos ^{2} 2 \alpha \sin ^{2} 2 \alpha+\sin ^{4} 2 \alpha= \\
& =\cos ^{4} 2 \alpha+\sin ^{4} 2 \alpha-6 \sin ^{2} 2 \alpha \cos ^{2} 2 \alpha= \\
& =\left(\cos ^{2} 2 \alpha+\sin ^{2} 2 \alpha\right)^{2}-2 \sin ^{2} 2 \alpha \cos ^{2} 2 \alpha-6 \sin ^{2} 2 \alpha \cos ^{2} 2 \alpha= \\
& =1-8 \sin ^{2} 2 \alpha \cos ^{2} 2 \alpha=1-2\left(4 \sin ^{2} 2 \alpha \cos ^{2} 2 \alpha\right)=1-2 \sin ^{2} 4 \alpha=\cos 8 \alpha
\end{aligned}
$$

Answer: $\cos 8 \alpha$.