By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that $ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$ By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that
$$ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$$
Here's what I've done so far (starting from after expansion):
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x\sin^4x) + \sin^6x$
$\cos^6x + (3\cos^2x\sin^2x)(\cos^2x+\sin^2x) + \sin^6x$
$\cos^6x + (3\cos^2x\sin^2x) + \sin^6x$
$\cos^6x + \sin^6x = -3\cos^2x\sin^2x$
$\cos^6x + \sin^6x = (-3/2)(2\cos^2x\sin^2x)$
$\cos^6x + \sin^6x = (-3/2)(\sin^22x)$
How can I get it into $ 1 - (3/4)\sin^2(2x)$?

You have nice answers by expansion, so I present a solution without it.
$$
\sin^6x + \cos^6x 
$$
$$
= \sin^6x + (1-\sin^2 x)^3
$$
$$
= \sin^6x + 1 - 3\sin^2 x+3\sin^4 x-\sin^6 x
$$
$$
= 1 - 3\sin^2 x + 3\sin^4 x 
$$
$$
= 1 - 3\sin^2(1 - \sin^2 x)
$$
$$
= 1 - 3\sin^2 x\cos^2 x
$$
$$
= 1 - \frac{3}{4}\sin^22x
$$Hence Proved.
