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

Dec 20, 2016

Start by factoring ${\sin}^{3} x - {\cos}^{3} x$ using the difference of cubes formula ${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$.

So,

$\frac{{\sin}^{3} x - {\cos}^{3} x}{\sin x - \cos x} = 1 + \sin x \cos x$

$\frac{\left(\sin x - \cos x\right) \left({\sin}^{2} x + \sin x \cos x + {\cos}^{2} x\right)}{\sin x - \cos x} = 1 + \sin x \cos x$

The $\sin x - \cos x$ cancel each other out.

${\sin}^{2} x + \sin x \cos x + {\cos}^{2} x = 1 + \sin x \cos x$

Use the identity ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$:

$1 + \sin x \cos x = 1 + \sin x \cos x$

$L H S = R H S$

Hopefully this helps!

Dec 20, 2016

See below.

#### Explanation:

From the polynomial identity

$\frac{{a}^{3} - {b}^{3}}{a - b} = {a}^{2} + a b + {b}^{2}$ we conclude that

$\frac{{\sin}^{3} x - {\cos}^{3} x}{\sin x - \cos x} = {\sin}^{2} x + \sin x \cos x + {\cos}^{2} x = 1 + \sin x \cos x$