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.

Consider $a^3+b^3=(a+b)^3-3ab(a+b)$ and $a^2+b^2=(a+b)^2-2ab$. For $a=\sin^2x$ and $b=\cos^2x$ we have
$$
2(\sin^6x+\cos^6x)-3(\sin^4x+\cos^4x)+1=
2(a+b)^3-6ab(a+b)-3(a+b)^2+6ab+1
$$
However, $a+b=\sin^2x+\cos^2x=1$, so the expression simplifies to
$$
2-6ab-3+6ab+1=0
$$
