<s>[INST] How to prove $\cos^4x - \sin^4x - \cos^2x + \sin^2x$ is always $0$? So I have a small problem here where I have to prove the following :

$$\cos^4x - \sin^4x - \cos^2x + \sin^2x = 0 $$

I know that the 2nd part is always $1$, so I need to prove that the first part also equals $1$. So how should I prove it ?
Edit : Sorry, the equation itself was wrong. I've edited it. 
 [/INST] Simple: it's not even true. Unless $x$ is a very specific value.
Your other statement, that's a different story. You can write down
$$\cos^4 x + \sin^4 x = \cos^4 x + \sin^4 x-2\sin^2 x \cos^2x + 2\sin^2x \cos^2 x =$$
$$= (\cos^2x+\sin^2x)^2-2\sin^2 x \cos^2 x = $$
$$=1-2\sin^2 x \cos^2 x=\cos^2 x + \sin^2 x \underbrace{(1-2\cos^2 x)}_{\neq \pm 1}$$
So not even this is true. Nothing's true here.

EDIT:
Now it's obvious.
$$\cos^4 x - \sin^4 x -\cos^2 x +\sin^2 x = $$
$$(\cos^2 x -\sin^2 x )\underbrace{(\cos^2 x + \sin^2 x )}_{1}-\cos^2 x +\sin^2 x = 0$$
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