# 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)

Jun 11, 2018

Reminder of algebraic identities:
${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} + a b + {b}^{2}\right)$
$f \left(x\right) = {\sin}^{6} x + {\cos}^{6} x$.
Call ${\sin}^{2} x = X$ and ${\cos}^{2} x = Y$ -->
$f \left(X\right) = {X}^{3} + {Y}^{3} = \left(X + Y\right) \left({X}^{2} + {Y}^{2} - X Y\right)$ (1)
Note.
$X + Y = {\sin}^{2} x + {\cos}^{2} x = 1$
$X Y = {\sin}^{2} x . {\cos}^{2} x = \left(\frac{1}{4}\right) {\sin}^{2} 2 x$
${X}^{2} + {Y}^{2} = {\left(X + Y\right)}^{2} - 2 X Y = 1 - \left(\frac{2}{4}\right) {\sin}^{2} 2 x$
Equation (1) becomes:
$f \left(x\right) = \left(1\right) \left(1 - \left(\frac{2}{4}\right) {\sin}^{2} 2 x - \left(\frac{1}{4}\right) {\sin}^{2} 2 x\right)$
Finally,
$f \left(x\right) = 1 - \left(\frac{3}{4}\right) {\sin}^{2} 2 x$. Proved