Question

How do you do (sin^4x - sin^2x cos^2x + cos^4x) ?

In fact, the problem was: sin^6x + cos^6x = 1-3/4 sen^2 2x
But... I factor the identities but I do not remember what else to do.
(sin^2x + cos^2x) (sin^4x - sen^2xcos^2x + cos^4x)

Answer 1

Reminder of algebraic identities:
`a^3 + b^3 = (a + b)(a^2 + ab + b^2)`
`f(x) = sin^6 x + cos^6 x`.
Call `sin^2 x = X` and `cos^2 x = Y` -->
`f(X) = X^3 + Y^3 = (X + Y)(X^2+ Y^2 - XY)` (1)
Note.
`X + Y = sin^2 x + cos^2 x = 1`
`XY = sin^2 x.cos ^2 x = (1/4)sin^2 2x`
`X^2 + Y^2 = (X + Y)^2 - 2XY = 1 - (2/4)sin^2 2x`
Equation (1) becomes:
`f(x) = (1)(1 - (2/4)sin^2 2x - (1/4)sin^2 2x)`
Finally,
`f(x) = 1 - (3/4)sin^2 2x`. Proved