3.297. $\sin ^{2}(2 \alpha-\beta)-\sin ^{2} 2 \alpha-\sin ^{2} \beta$.
## Solution.

$$
\sin ^{2}(2 \alpha-\beta)-\sin ^{2} 2 \alpha-\sin ^{2} \beta=\left[\sin ^{2} \frac{x}{2}=\frac{1-\cos x}{2}\right]=
$$

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

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