Question

sin cube theta plus cos cube theta by sin theta + cos theta=​

Answer 1

$\frac{ {sin}^{3} \theta + {cos}^{3} \theta}{sin \theta + cos \theta} \\ \\ = \frac{(sin \theta + cos \theta)( {sin}^{2} \theta - sin \theta \: cos \theta + {cos}^{2} \theta )}{sin \theta + cos \theta} \\ \\ = 1 - sin \theta cos \theta$