Question

Please prove the following identity

Answer 1

Let's remember the factorizations

$a^3 + b^3 = (a+b)(a^2-ab+b^2)$

$a^3- b^3=(a-b)(a^2+ab+b^2)$

Now

$\dfrac{\sin ^3 \theta + \cos^3 \theta}{\sin \theta + \cos \theta} + \dfrac{\sin^3 \theta - \cos^3 \theta}{\sin \theta - \cos \theta}$

$=\dfrac{(\sin \theta + \cos \theta)(\sin ^2 \theta - \sin \theta\cos \theta + \cos^2 \theta)}{\sin \theta + \cos \theta} + \dfrac{(\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)}{\sin \theta - \cos \theta}$

$=\sin ^2 \theta - \sin \theta\cos \theta + \cos^2 \theta + \sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta$

$=\sin ^2 \theta + \cos^2 \theta + \sin^2 \theta+ \cos^2 \theta$

$=2 \quad\checkmark$