Show $\cos(x+y)\cos(x-y) - \sin(x+y)\sin(x-y) = \cos^2x - \sin^2x$ 
Show $\cos(x+y)\cos(x-y) - \sin(x+y)\sin(x-y) = \cos^2x - \sin^2x$

I have got as far as showing that:
$\cos(x+y)\cos(x-y) = \cos^2x\cos^2y -\sin^2x\sin^2y$
and 
$\sin(x+y)\sin(x-y) = \sin^2x\cos^2y - \cos^2x\sin^2y$
I get stuck at showing:
$\cos^2x\cos^2y -\sin^2x\sin^2y - \sin^2x\cos^2y - \cos^2x\sin^2y = \cos^2x - \sin^2x$
I know that $\sin^2x + \cos^2x = 1$ and I have tried rearranging this identity in various ways, but this has not helped me so far.

Hint:     $$\cos(a+b)=\cos a \cos b-\sin a \sin b$$$$\cos(2a)=\cos^2 a -\sin ^2a $$

$$\begin{align}\cos(x+y)\cos(x-y) - \sin(x+y)\sin(x-y) &= \cos((x+y)+(x-y))\\&=\cos2x\\&=\cos^2x - \sin^2x\end{align}$$

