Use the sum of cubes factoring rule
[tex]a^3 + b^3 = (a+b)(a^2 - ab + b^2)[/tex]
to transform the left hand side into the right hand side.
[tex]\frac { \sin^{3} \theta + \cos^{3} \theta } { \sin \theta + \cos \theta } = 1 - \sin \theta \cdot \cos \theta\\\\\frac { (\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cdot \cos\theta + \cos^2 \theta) } { \sin \theta + \cos \theta } = 1 - \sin \theta \cdot \cos \theta\\\\[/tex]
[tex]\sin^2 \theta - \sin \theta \cdot \cos\theta + \cos^2 \theta = 1 - \sin \theta \cdot \cos \theta\\\\(\sin^2 \theta + \cos^2\theta)- \sin \theta \cdot \cos\theta = 1 - \sin \theta \cdot \cos \theta\\\\1- \sin \theta \cdot \cos\theta = 1 - \sin \theta \cdot \cos \theta \ \ \checkmark\\\\[/tex]
Throughout the entire process, the right hand side stayed the same.
On the last step, I used the pythagorean identity.
