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

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

$$
\begin{aligned}
& =\cos ^{3} \alpha-\sin ^{3} \alpha+\sin \alpha-\cos \alpha= \\
& =(\cos \alpha-\sin \alpha)\left(\cos ^{2} \alpha+\sin \alpha \cos \alpha+\sin ^{2} \alpha\right)-(\cos \alpha-\sin \alpha)= \\
& =(\cos \alpha-\sin \alpha)\left(\cos ^{2} \alpha+\sin \alpha \cos \alpha+\sin ^{2} \alpha-1\right)= \\
& =(\cos \alpha-\sin \alpha)(1+\sin \alpha \cos \alpha-1)=(\cos \alpha-\sin \alpha) \sin \alpha \cos \alpha= \\
& =\frac{2}{\sqrt{2}}\left(\frac{\sqrt{2}}{2} \cos \alpha-\frac{\sqrt{2}}{2} \sin \alpha\right) \cdot \frac{2 \sin \alpha \cos \alpha}{2}= \\
& =\frac{\sqrt{2}}{2}\left(\cos \frac{\pi}{4} \cos \alpha-\sin \frac{\pi}{4} \sin \alpha\right) \sin 2 \alpha= \\
& =[\cos x \cos y-\sin x \sin y=\cos (x+y)]=\frac{\sqrt{2}}{2} \cos \left(\frac{\pi}{4}+\alpha\right) \sin 2 \alpha
\end{aligned}
$$

Answer: $\frac{\sqrt{2}}{2} \sin 2 \alpha \cos \left(\frac{\pi}{4}+\alpha\right)$.