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.
 A: $$\sin^4 x+\cos^4 x=(\sin^2 x+\cos^2 x)^2-2\sin^2x\cos^2x=1-2\sin^2x\cos^2x$$
$$\sin^6 x+\cos^6 x=(\sin^2 x+\cos^2 x)^3-3\sin^2 x\cos^2 x(\sin^2 x+\cos^2 x)=1-3\sin^2x\cos^2x$$
Thus your expression simplifies to
$$2-6\sin^2x\cos^2x-3+6\sin^2x\cos^2x+1$$
Which is $0$, for all $x$.
A: $$2(\sin^6x +\cos^6x) - 3(\sin^4x+\cos^4x)+1=$$
$$=2(\sin^2x +\cos^2x)(\sin^4x-\sin^2x \cos^2 x+\cos^4x) - 3(\sin^4x+\cos^4x)+1=$$
$$=2(\sin^4x-\sin^2x \cos^2 x+\cos^4x) - 3(\sin^4x+\cos^4x)+1=$$
$$=-2\sin^2x \cos^2 x -\sin^4x-\cos^4x+1=$$
$$-(\sin^2x +\cos^2x)^2+1=-1+1=0$$
A: Let $t=\sin^2(x)$.
$$2(t^3+(1-t)^3)-3(t^2+(1-t)^2)+1=6t^2-6t+2-6t^2+6t-3+1=0.$$
A: $$
\begin{align}
1
&=\left(\sin^2(x)+\cos^2(x)\right)^3\\
&=\sin^6(x)+3\sin^4(x)\cos^2(x)+3\sin^2(x)\cos^4(x)+\cos^6(x)\\
&=\sin^6(x)+3\sin^2(x)\cos^2(x)+\cos^6(x)\tag{1}\\\\
1
&=\left(\sin^2(x)+\cos^2(x)\right)^2\\
&=\sin^4(x)+2\sin^2(x)\cos^2(x)+\cos^4(x)\tag{2}
\end{align}
$$
Subtracting $3$ times $(2)$ from $2$ times $(1)$ gives $-1$. Therefore,
$$
2\left(\sin^6(x)+\cos^6(x)\right)-3\left(\sin^4(x)+\cos^4(x)\right)+1=0\tag{3}
$$
A: 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
$$
2-6ab-3+6ab+1=0
$$
