3.254. $4 \cos \left(\alpha-\frac{\pi}{2}\right) \sin ^{3}\left(\frac{\pi}{2}+\alpha\right)-4 \sin \left(\frac{5}{2} \pi-\alpha\right) \cos ^{3}\left(\frac{3}{2} \pi+\alpha\right)$.
Solution.

$$
\begin{aligned}
& 4 \cos \left(\alpha-\frac{\pi}{2}\right) \sin ^{3}\left(\frac{\pi}{2}+\alpha\right)-4 \sin \left(\frac{5}{2} \pi-\alpha\right) \cos ^{3}\left(\frac{3}{2} \pi+\alpha\right)= \\
& =4 \cos \left(\frac{\pi}{2}-\alpha\right)\left(\sin \left(\frac{\pi}{2}+\alpha\right)\right)^{3}-4 \sin \left(\frac{5}{2} \pi-\alpha\right)\left(\cos \left(\frac{3}{2} \pi+\alpha\right)\right)^{3}= \\
& =4 \sin \alpha \cos ^{3} \alpha-4 \cos \alpha \sin ^{3} \alpha=4 \sin \alpha \cos \alpha\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right)=
\end{aligned}
$$

$=2(2 \sin \alpha \cos \alpha) \cos 2 \alpha=2 \sin 2 \alpha \cos 2 \alpha=\sin 4 \alpha$.

Answer: $\sin 4 \alpha$.