<s>[INST] 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.
 [/INST] $$
\begin{align}
1
&=\left(\sin^2(x)+\cos^2(x)\right)^3\\
&=\sin^6(x)+3\sin^4(x)\cos^2(x)+3\sin^2(x)\cos^4(x)+\cos^6(x)\\
&=\sin^6(x)+3\sin^2(x)\cos^2(x)+\cos^6(x)\tag{1}\\\\
1
&=\left(\sin^2(x)+\cos^2(x)\right)^2\\
&=\sin^4(x)+2\sin^2(x)\cos^2(x)+\cos^4(x)\tag{2}
\end{align}
$$
Subtracting $3$ times $(2)$ from $2$ times $(1)$ gives $-1$. Therefore,
$$
2\left(\sin^6(x)+\cos^6(x)\right)-3\left(\sin^4(x)+\cos^4(x)\right)+1=0\tag{3}
$$
</s>