# How do you verify (sin^3x+cos^3x)/(sinx+cosx)=1-sinxcosx?

Apr 22, 2016

Expansion of a cubic ${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$
$\frac{{\sin}^{3} x + {\cos}^{3} x}{\sin x + \cos x} = \frac{\left(\sin x + \cos x\right) \left({\sin}^{2} x - \sin x \cos x + {\cos}^{2} x\right)}{\sin x + \cos x}$
$= {\sin}^{2} x - \sin x \cos x + {\cos}^{2} x$
Identity: ${\sin}^{2} x + {\cos}^{2} x = 1$
$= {\sin}^{2} x + {\cos}^{2} x - \sin x \cos x$
$= 1 - \sin x \cos x$