30. If $\sin \theta-\cos \theta=\frac{\sqrt{6}-\sqrt{2}}{2}$, find the value of $24\left(\sin ^{3} \theta-\cos ^{3} \theta\right)^{2}$,
30. Ans: 12
$$
\begin{aligned}
\sin ^{3} \theta-\cos ^{3} \theta & =(\sin \theta-\cos \theta)\left(\sin ^{2} \theta+\sin \theta \cos \theta+\cos ^{2} \theta\right) \\
& =\frac{\sqrt{6}-\sqrt{2}}{2}(1+\sin \theta \cos \theta) .
\end{aligned}
$$

Now we have
$$
\begin{aligned}
(\sin \theta-\cos \theta)^{2}=\left(\frac{\sqrt{6}-\sqrt{2}}{2}\right)^{2} & \Longrightarrow 1-2 \sin \theta \cos \theta=\frac{6+2-2 \sqrt{12}}{4}=2-\sqrt{3} \\
& \Longrightarrow \sin \theta \cos \theta=\frac{\sqrt{3}-1}{2} .
\end{aligned}
$$

Hence we have
$$
\begin{aligned}
\sin ^{3} \theta-\cos ^{3} \theta & =\frac{\sqrt{6}-\sqrt{2}}{2}\left(1+\frac{\sqrt{3}-1}{2}\right) \\
& =\frac{\sqrt{2}(\sqrt{3}-1)}{2}\left(\frac{\sqrt{3}+1}{2}\right) \\
& =\frac{\sqrt{2}(3-1)}{4}=\frac{\sqrt{2}}{2} .
\end{aligned}
$$

Therefore,
$$
24\left(\sin ^{3} \theta-\cos ^{3} \theta\right)^{2}=24\left(\frac{\sqrt{2}}{2}\right)^{2}=24 \times \frac{1}{2}=12 .
$$