First of all, we should understand how to use the expansion of this formula: (a+b) 3 = a 3+3a 2b+3b 2a+b 3.

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