2. Calculate:
$$
\sin ^{3} 20^{\circ}-\sin ^{3} 100^{\circ}+\sin ^{3} 140^{\circ}=
$$
2. $-\frac{3 \sqrt{3}}{8}$.

From the triple angle formula, we get
$$
\begin{array}{l}
\sin ^{3} \alpha+\sin ^{3}\left(\alpha-120^{\circ}\right)+\sin ^{3}\left(\alpha+120^{\circ}\right) \\
=\frac{1}{4}\left[3 \sin \alpha-\sin 3 \alpha+3 \sin \left(\alpha-120^{\circ}\right)-\right. \\
\left.\quad \sin 3 \alpha+3 \sin \left(\alpha+120^{\circ}\right)-\sin 3 \alpha\right] \\
=-\frac{3}{4} \sin 3 \alpha .
\end{array}
$$

Then $\sin ^{3} 20^{\circ}-\sin ^{3} 100^{\circ}+\sin ^{3} 140^{\circ}$
$$
=-\frac{3}{4} \sin 120^{\circ}=-\frac{3 \sqrt{3}}{8} .
$$