12. Given $\sin \alpha-\cos \alpha=\frac{1}{2}$. Then the value of $\sin ^{3} \alpha-$ $\cos ^{3} \alpha$ is $\qquad$.
$$
\begin{array}{l}
\text { 12. } \frac{11}{16} \\
\sin ^{3} \alpha-\cos ^{3} \alpha \\
=(\sin \alpha-\cos \alpha)\left(\sin ^{2} \alpha+\sin \alpha \cdot \cos \alpha+\cos ^{2} \alpha\right) \\
=(\sin \alpha-\cos \alpha)(1+\sin \alpha \cdot \cos \alpha)
\end{array}
$$

Given $\sin \alpha-\cos \alpha=\frac{1}{2}$, and
$$
(\sin \alpha-\cos \alpha)^{2}=\sin ^{2} \alpha-2 \sin \alpha \cdot \cos \alpha+\cos ^{2} \alpha=\frac{1}{4} \text {, }
$$

thus $\sin \alpha \cdot \cos \alpha=\frac{3}{8}$.
Therefore, $\sin ^{3} \alpha-\cos ^{3} \alpha=\frac{11}{16}$.