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)

1 Answer
Jun 11, 2018

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