# How do you verify [sin^3(B) + cos^3(B)] / [sin(B) + cos(B)] = 1-sin(B)cos(B)?

Expansion of ${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$, and we can use this:
$\frac{{\sin}^{3} B + {\cos}^{3} B}{\sin B + \cos B} = \frac{\left(\sin B + \cos B\right) \left({\sin}^{2} B - \sin B \cos B + {\cos}^{2} B\right)}{\sin B + \cos B}$
$= {\sin}^{2} B - \sin B \cos B + {\cos}^{2} B$
$= {\sin}^{2} B + {\cos}^{2} B - \sin B \cos B$ (identity: ${\sin}^{2} x + {\cos}^{2} x = 1$)
$= 1 - \sin B \cos B$