Simplify
\[\frac{\sin^4 x + \cos^4 x - 1}{\sin^6 x + \cos^6 x - 1}.\]
I try to rewrite the numerator as a difference of squares: $\sin^4 x + \cos^4 x - 1 = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x - 1$. Now I can use the identity and simplify: $(\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x - 1 = 1 - 2\sin^2 x \cos^2 x - 1 = -2\sin^2 x \cos^2 x$. I do the same thing for the denominator: $\sin^6 x + \cos^6 x - 1 = (\sin^2 x + \cos^2 x)^3 - 3\sin^4 x \cos^2 x - 3\sin^2 x \cos^4 x - 1$. Again, I use the identity and simplify: $(\sin^2 x + \cos^2 x)^3 - 3\sin^4 x \cos^2 x - 3\sin^2 x \cos^4 x - 1 = 1 - 3\sin^4 x \cos^2 x - 3\sin^2 x \cos^4 x - 1 = -3\sin^4 x \cos^2 x - 3\sin^2 x \cos^4 x$. Now I have $\frac{\sin^4 x + \cos^4 x - 1}{\sin^6 x + \cos^6 x - 1} = \frac{-2\sin^2 x \cos^2 x}{-3\sin^4 x \cos^2 x - 3\sin^2 x \cos^4 x}$. I can simplify this further by canceling out a factor of $-1$ and a factor of $\sin^2 x \cos^2 x$: $\frac{-2\sin^2 x \cos^2 x}{-3\sin^4 x \cos^2 x - 3\sin^2 x \cos^4 x} = \frac{2}{3\sin^2 x + 3\cos^2 x}$. Finally, I can use the identity one more time and simplify: $\frac{2}{3\sin^2 x + 3\cos^2 x} = \frac{2}{3(\sin^2 x + \cos^2 x)} = \frac{2}{3}$. This is the simplest form of the expression. 