To simplify the expression \( \cos^4 x - \sin^4 x \), we can use the identity for the difference of squares:

\[
a^2 - b^2 = (a - b)(a + b)
\]

In this case, let \( a = \cos^2 x \) and \( b = \sin^2 x \). Then we have:

\[
\cos^4 x - \sin^4 x = (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x)
\]

Since \( \cos^2 x + \sin^2 x = 1 \), the expression simplifies to:

\[
\cos^4 x - \sin^4 x = (\cos^2 x - \sin^2 x)(1) = \cos^2 x - \sin^2 x
\]

Next, let's verify the second part of your expression.

You've written:

\[
1 - 2\sin^2 x \cos^4 x - \sin^4 x = 1 - 2\sin^2 x
\]

It seems there might be an error in relating the expressions or making an algebraic mistake. To analyze it further, we can rewrite \( \cos^4 x \) in terms of \( \sin^2 x \):

Since \( \cos^2 x = 1 - \sin^2 x \), we can rewrite \( \cos^4 x \) as:

\[
\cos^4 x = (1 - \sin^2 x)^2 = 1 - 2\sin^2 x + \sin^4 x
\]

Now substituting \( \cos^4 x \) into your expression gives:

\[
\cos^4 x - \sin^4 x = (1 - 2\sin^2 x + \sin^4 x) - \sin^4 x = 1 - 2\sin^2 x
\]

We find that:

\[
\cos^4 x - \sin^4 x = 1 - 2\sin^2 x
\]

Thus the simplifications hold true:

\[
\cos^4 x - \sin^4 x = 1 - 2\sin^2 x
\]

So, to summarize:

\[
\cos^4 x - \sin^4 x = \cos^2 x - \sin^2 x = 1 - 2\sin^2 x
\]

Both sides of the equation you provided are equal and valid. If you have more specific questions or further expressions to analyze, feel free to share!