algebra precalculus - A cubed trigonometric identity?

Could somebody please show why the following is a trigonometric identity?

$$\dfrac{\sin^3 a - \cos^3a}{\sin a - \cos a} = 1 + \sin a \cos a$$

This problem appears on page $48$ of Gelfand's and Saul's "Trigonometry". (It's not homework.)

It is probably the fact that we are dealing with trig ratios cubed that is throwing me off.

A question with squared trig ratios usually gives me no troubles.

I keep running into a mess. For example: I've multiplied the numerator and denominator by $\sin a + \cos a$ with no luck; and likewise, by $\sin a - \cos a$.