3.296. $\sin ^{2}(\alpha-2 \beta)-\cos ^{2} \alpha-\cos ^{2} 2 \beta$.

3.296. $\sin ^{2}(\alpha-2 \beta)-\cos ^{2} \alpha-\cos ^{2} 2 \beta$.
Solution.

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

Answer: $-2 \cos \alpha \cos 2 \beta \cos (\alpha-2 \beta)$.