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)

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)

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Nghi N. Share
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

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