Prove that 4(sin^6(x) + cos^6(x)) = 4 - 3sin^2(2x)?

Question

How do you prove that 4(sin^6(x) + cos^6(x)) = 4 - 3sin^2(2x)?

My teacher said something about putting sin^6(x) = (sin^2(x))^3, but that didn't really help me...

HELP!

Note: I am supposed to prove this without using a calculator. I am supposed to prove this using algebra and trigonometric formulas (ie. double angle, half angle, reduction of powers, etc.)

Answer 1

use a tech calculator with all the signs on it most of them are made by texasinstraments they can give you the answer you need for theis question.

Answer 2

4(sin^6(x) + cos^6(x))
=4((sin^2(x))^3+(cos^2(x))^3)
use the formula
a^3+b^3=(a+b)^3-3ab(a+b)
=4(sin^2(x)+cos^2(x))^3-
3sin^2(x)cos^2(x){sin^2(x)+cos^2(x)}
=4(1)-12sin^2(x)cos^2(x)
=4 -3{2sinxcosx}^2
=4 - 3sin^2(2x)