Question

Prove that

${\sin }^3\theta +{\cos }^3\theta$ = $(1-\sin \theta .\cos \theta )$($\sin \theta +\cos \theta$)

Answer 1

$a^3+b^3=(a+b)(a^2+b^2-ab)$
${\sin }^3\theta +{\cos }^3\theta =\left(\sin \theta +\cos \theta \right)\left({\sin }^2\theta -\sin \theta .\cos \theta +{\cos }^2\theta \right)$
We know that ${\sin }^2\theta +{\cos }^2\theta =1$
${\sin }^3\theta +{\cos }^3\theta =\left(\sin \theta +\cos \theta \right)\left(1-\sin \theta .\cos \theta \right)$