Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

$\dfrac{\text{sin}^3 A + \text{cos}^3 A}{\text{sin A + cos A}} + \dfrac{\text{sin}^3 A - \text{cos}^3 A}{\text{sin A - cos A}} = 2$.

We know that,

a³ + b³ = (a + b)(a² - ab + b²).

and

a³ - b³ = (a - b)(a² + ab + b²).

Using above formulas, the L.H.S. of the equation can be written as,

$\Rightarrow \dfrac{\text{(sin A + cos A)(sin}^2 A - \text{sin A cos A + cos}^2 A)}{\text{sin A + cos A}} + \dfrac{\text{(sin A - cos A)(sin}^2 A + \text{sin A cos A + cos}^2 A)}{\text{sin A - cos A}} \\[1em] \Rightarrow \dfrac{\cancel{\text{(sin A + cos A)}}\text{(sin}^2 A - \text{sin A cos A + cos}^2 A)}{\cancel{\text{(sin A + cos A)}}} + \dfrac{\cancel{\text{(sin A - cos A)}} \text{(sin}^2 A + \text{sin A cos A + cos}^2 A)}{\cancel{\text{(sin A - cos A)}}} \\[1em] \Rightarrow \text{sin}^2 A + \text{cos}^2 A - \text{sin A cos A} + \text{sin}^2 A + \text{cos}^2 A + \text{sin A cos A} \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2.$

Since, L.H.S. = R.H.S. hence proved that, $\dfrac{\text{sin}^3 A + \text{cos}^3 A}{\text{sin A + cos A}} + \dfrac{\text{sin}^3 A - \text{cos}^3 A}{\text{sin A - cos A}} = 2$.