# 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.
You can factor $x^3 - y^3$ as $(x - y)(x^2 + xy + y^2)$. So divide by $(x - y)$. Then,
\begin{align}\frac{\sin(a)^3 - \cos(a)^3}{\sin(a) - \cos(a)} &= \sin(a)^2 + \sin(a) \cos(a) + \cos(a)^2\\ &= \sin(a)^2 + \cos(a)^2 + \sin(a) \cos(a)\\ &= 1 + \sin(a) \cos(a)\end{align}.