Simplify $2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$ Here is the expression:

$$2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$$

The exercise is to evaluate it. 
In my text book the answer is $0$
I tried to factor the expression, but it got me nowhere.

$$2(\sin^6x +\cos^6x) - 3(\sin^4x+\cos^4x)+1=$$
$$=2(\sin^2x +\cos^2x)(\sin^4x-\sin^2x \cos^2 x+\cos^4x) - 3(\sin^4x+\cos^4x)+1=$$
$$=2(\sin^4x-\sin^2x \cos^2 x+\cos^4x) - 3(\sin^4x+\cos^4x)+1=$$
$$=-2\sin^2x \cos^2 x -\sin^4x-\cos^4x+1=$$
$$-(\sin^2x +\cos^2x)^2+1=-1+1=0$$
