You need to move all the terms that contain `sin x` and `cos x` to the left side, such that:

`sin^4 x + 2sin^2 x*cos^2 x + cos^4 x = 1`

You need to use the following special product, such that:

`(a + b)^2 = a^2 + 2ab + b^2`

Considering `a = sin^2 x` and `b = cos^2 x` , yields:

`sin^4 x + 2sin^2 x*cos^2 x + cos^4 x = (sin^2 x + cos^2 x)^2`

Using the Pythagorean trigonometric identity, `sin^2 x + cos^2 x = 1` , yields:

`sin^4 x + 2sin^2 x*cos^2 x + cos^4 x =1^2 = 1`

**Hence, the last line proves that the given identity `sin^4 x + cos^4 x = 1 - 2sin^2 x*cos^2 x` holds.**

There are a couple of ways to show this. I prefer to get both sides to be the same thing. For instance, the left side can be written as:

(**(sin x)^2 + (cos x)^2**)^2 - 2(sin x)^2 (cos x)^2

Well, the bold part of the parenthesis is equal to 1. So:

1 - 2(sin x)^2 (cos x)^2 = 1 - 2 (sin x cos x)^2

We can combine the squares on the left for:

1 - 2 (sin x cos x)^2 = 1 - 2 (sin x cos x)^2

Thus, we have equality.