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 $\sin a - \cos a$.

Recall that
$$(x^3 - y^3) = (x-y)(x^2 + y^2 + xy)$$ Hence, $$\dfrac{x^3 - y^3}{x-y} = x^2 + y^2 + xy$$
Use the above identity and make use of the fact that $\sin^2(\theta) + \cos^2(\theta) = 1$, to get what you want.
