Take (cos 2x) as a whole, which is equivalent to A (sin 2 x) as a whole, which is equivalent to B.
The solution to this problem
Prove:
(cos^6 X)+(sin^6 X)
= [(cos 2 ten)] 3 ]^3+ [(sin^2 x)]^3)] 3
=[(cos^2 x)]+[(sin^2 x)]*[(cos^4 x )-[(cos^2 x )(sin^2 x)+[(sin^4x)]
= 1 * {[(cos^2 x)+(sin^2 x)]^2-3(cos^2x)(sin^2 x)}
=1-3 (cos 2x) (sin 2 x)
=1-3 (1-3sin 2x) (sin 2 x)
= 1 -3sin^2 x + 3sin^4 x