A geometry problem by Prince Raj

Bom! para resolver o problema proposto, fazemos uso da fatoração entre diferença entre cubos e a relação fundamental da trigonometria, veja:

$\mathsf{\sin^3 x - \cos^3 x =} \\\\ \mathsf{(\sin x - \cos x) \cdot (\sin^2 x + \sin x \cdot \cos x + \cos^2 x)} = \\\\ \mathsf{(\sin x - \cos x) \cdot (\underbrace{\mathsf{\sin^2 x + \cos^2 x}}_{\mathsf{\sin^2 x + \cos^2 x = 1}} + \sin x \cdot \cos x) =}$

$\boxed{\mathsf{(\sin x - \cos x) \cdot (1 + \sin x \cdot \cos x)}}$

A^3 - B^3 factors as ( A - B ) (A^2 + A B + B^2 )

So (sin x - cos x) * ( sin^2 x + sin x * cos x + cos^2 x )

But sin^2 x + cos^2 x = 1

(sin x - cos x) * ( 1 + sin x * cos x )