3.186. $\frac{\sin ^{4} \alpha+\cos ^{4} \alpha-1}{\sin ^{6} \alpha+\cos ^{6} \alpha-1}=\frac{2}{3}$.
Solution.

$$
\begin{aligned}
& \frac{\sin ^{4} \alpha+\cos ^{4} \alpha-1}{\sin ^{6} \alpha+\cos ^{6} \alpha-1}= \\
& =\frac{\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)^{2}-2 \sin ^{2} \alpha \cos ^{2} \alpha-1}{\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)\left(\sin ^{4} \alpha-\sin ^{2} \alpha \cos ^{2} \alpha+\cos ^{4} \alpha\right)-1}= \\
& =\frac{1-2 \sin ^{2} \alpha \cos ^{2} \alpha-1}{\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)^{2}-3 \sin ^{2} \alpha \cos ^{2} \alpha-1}=\frac{-2 \sin ^{2} \alpha \cos ^{2} \alpha}{1-3 \sin ^{2} \alpha \cos ^{2} \alpha-1}= \\
& =\frac{-2 \sin ^{2} \alpha \cos ^{2} \alpha}{-3 \sin ^{2} \alpha \cos ^{2} \alpha}=\frac{2}{3}
\end{aligned}
$$

The identity is proven.