Simplifying $(\sin^6x + \cos^6x) - (\sin^4x + \cos^4x) + \sin^2x\cos^2x$

Question

$(\sin^6x + \cos^6x) - (\sin^4x + \cos^4x) + \sin^2x\cos^2x =$

Right Answer: $0$, but I could not solve this question. Help me please.

Answer 1

$$(\sin^6x + \cos^6x) - (\sin^4x + \cos^4x) + \sin^2x\cos^2x $$ $$= \sin^6x + \cos^6x - \sin^4x -\cos^4x+\sin^2x\cos^2x$$ $$= \cos^6x(\tan^6x + 1 - \tan^4x\sec^2{x} -\sec^2x+\tan^2x\sec^2x)$$ $$= \cos^6x(\tan^6x + 1 - \tan^4x(1+\tan^2x) -(1+\tan^2x)+\tan^2x(1+\tan^2x))$$ $$= \cos^6x(\tan^6x + 1 - \tan^4x-\tan^6x -1-\tan^2x+\tan^2x+\tan^4x)$$ $$= \cos^6x(0)$$ $$= 0$$

Answer 2

hint: $\sin^2 x + \cos^2 x = 1$; if you square that, you get $$ \sin^4 x + 2\sin^2 x \cos^2 x + \cos^4 x = 1, $$ which you can turn into $$ \sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x $$ and that lets you get rid of the 4th powers in the left hand side.

Can you run with that idea?

Answer 3

With $c:=\cos^2 x,\,s:=\sin^2 x$, it suffices to note $$c^3+s^3-c^2-s^2+cs=(c+s-1)(c^2-cs+s^2).$$