How can I prove that $2(\cos^6(x)-\sin^6(x))-3(\cos^4(x)+\sin^4(x))=-4\sin^6(x)-1$ How can I prove that $2(\cos^6(x)-\sin^6(x))-3(\cos^4(x)+\sin^4(x))=-4\sin^6(x)-1$
I tried to factor and I got $2\cos^4(x)+(-2\sin^2(x)-3)(\cos^4(x)+\sin^4(x))$ but that doesn't lead me to my goal.
I also tried to write all the cosines in terms of sines  to have: $-3\sin^6(x)-9\sin^2(x)-2-3\sin^4(x)$
But I don't see how to continue
Any hint is welcome! thnxx 

According to the equation:
$$2(cos^6(x) - sin^6(x)) - 3(cos^4(x) + sin^4(x)) = -4sin^6(x) - 1$$
this means that:
$$\frac{2(cos^6(x) - sin^6(x)) - 3(cos^4(x) + sin^4(x)) + 1}{-4} = sin^6(x)$$
And since we are dividing by a negative where $x\in \mathbb{Q}$ I assume, this means that:
$$3(cos^4(x) + sin^4(x)) - 1 > 2(cos^6(x) - sin^6(x))$$
Meaning that:
$$\frac{3(cos^4(x) + sin^4(x)) - 1}{2} - cos^6(x) + 2sin^6(x) > sin^6(x) = \frac{2(cos^6(x) - sin^6(x)) - 3(cos^4(x) + sin^4(x)) + 1}{-4}$$
Therefore when we do some of the simplifications, we find that:
$$-4sin^6(x) - 1 < 2 - 6(cos^4(x) + sin^4(x)) + 4cos^6(x) + 8sin^6(x)$$
$$\implies sin^6(x) < \frac{3 - 6(cos^4(x) + sin^4(x)) + 4cos^6(x) + 8sin^6(x)}{-4}$$
$$\therefore \frac{3(cos^4(x) + sin^4(x)) - 1}{2} - cos^6(x) + 2sin^6(x) = \frac{3 - 6(cos^4(x) + sin^4(x)) + 4cos^6(x) + 8sin^6(x)}{-4}$$
So there's your proof. Since there is no contradiction and we basically come to a fact something like $-4n = \frac{-4n}{-4}$ then we cannot go any further with this, meaning that every single equation in this answer is true! So there ya go. Hope I helped! :)
