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.

$$\sin^4 x+\cos^4 x=(\sin^2 x+\cos^2 x)^2-2\sin^2x\cos^2x=1-2\sin^2x\cos^2x$$
$$\sin^6 x+\cos^6 x=(\sin^2 x+\cos^2 x)^3-3\sin^2 x\cos^2 x(\sin^2 x+\cos^2 x)=1-3\sin^2x\cos^2x$$
Thus your expression simplifies to
$$2-6\sin^2x\cos^2x-3+6\sin^2x\cos^2x+1$$
Which is $0$, for all $x$.
