Verify $\frac{\sin^3A + \cos^3A}{\sin A + \cos A} = 1 - \sin A\cos A$ How can I verify the following trigonometric identity?
$$\frac{\sin^3 A + \cos^3 A}{\sin A+\cos A} = 1-\sin A\cos A.$$
My work so far is 
$$\begin{align*}
&\frac{\sin\cos(\sin^2+\cos^2)}{\sin+\cos}\\
&\frac{\sin\cos(1)}{\sin+\cos}
\end{align*}$$
 A: First: don't use $\sin$ and $\cos$ without arguments. For instance, your last formula, it's easy to get confused and think you are computing, inter alia, $\cos(1)$, which you are not.
Second: it seems that you think that
$$\sin^3 A + \cos^3A \text{ is equal to } \sin A\cos A(\sin^2A + \cos^2A).$$
They are not equal; you can verify that by actually multiplying out the right hand side; you will see that you get
$$\sin^3A\cos A + \cos^3A\sin A\neq \sin^3A + \cos^3A.$$
There are a few algebraic formulas that show up a lot and it is good to know them. When I was in middle school, they were known as "notable products", because they were, well, notable and showed up a lot. One was the square of a binomial, $(a+b)^2 = a^2+2ab+b^2$; one was the "difference of squares" or "conjugate product": $(a+b)(a-b)=a^2-b^2$.
Two others are the difference-of-cubes and the sum-of-cubes:
$$\begin{align*}
a^3-b^3 &= (a-b)(a^2+ab+b^2)\\
a^3+b^3 &= (a+b)(a^2-ab+b^2).
\end{align*}$$
Using the second one, you can simplify the left hand side,
$$\frac{\sin^3 A + \cos^3 A}{\sin A + \cos A}$$
by letting $a=\sin A$ and $b=\cos A$, and proceed from there.
A: $$
a^3+b^3 = (a+b)(a^2-ab+b^2).
$$
Use that to factor the numerator.
Also, it is not correct to say that $s^3+c^3 = sc(s^2+c^2)$.
