Trigonometry
Prove the identity $\;$ $\dfrac{\sin^3 A - \cos^3 A}{1 + \sin A \cos A} = \sin A - \cos A$
$\begin{aligned}
LHS & = \dfrac{\sin^3 A - \cos^3 A}{1 + \sin A\cos A} \\\\
& \left[\text{Note: } a^3 - b^3 = \left(a - b\right) \left(a^2 + ab + b^2\right)\right] \\\\
& = \dfrac{\left(\sin A - \cos A\right) \left(\sin^2 A + \sin A \cos A + \cos^2 A\right)}{1 + \sin A \cos A} \\\\
& \left[\text{Note: }\sin^2 A + \cos^2 A = 1\right] \\\\
& = \dfrac{\left(\sin A - \cos A\right) \left(1 + \sin A \cos A\right)}{1 + \sin A \cos A} \\\\
& = \sin A - \cos A = RHS
\end{aligned}$
Hence proved.