25. If $\cos (2 A)=-\frac{\sqrt{5}}{3}$, find the value of $6 \sin ^{6} A+6 \cos ^{6} A$.
25. Answer: 4
Using the identity $\left(a^{3}+b^{3}\right)=(a+b)\left(a^{2}-a b+b^{2}\right)$ with $a=\sin ^{2} A$ and $b=\cos ^{2} A$, we have
$$
\begin{aligned}
\sin ^{6} A+\cos ^{6} A & =\left(\sin ^{2} A+\cos ^{2} A\right)\left(\left(\sin ^{2} A+\cos ^{2} A\right)^{2}-3 \sin ^{2} A \cos ^{2} A\right) \\
& =1-3 \sin ^{2} A \cos ^{2} A=1-\frac{3}{4} \sin ^{2} 2 A \\
& =1-\frac{3}{4}\left(1-\cos ^{2} 2 A\right)=1-\frac{3}{4}\left(1-\frac{5}{9}\right)=\frac{2}{3} .
\end{aligned}
$$

Hence $6 \sin ^{6} A+6 \cos ^{6} A=4$.