Question

prove that sin^3+cos^3/cos+sin +sin^3-cos^3/sin-cos= 2

Answer 1

$\frac{ {sin}^{3}θ + {cos}^{3}θ }{cosθ + sinθ} + \frac{ {sin}^{3}θ - {cos}^{3}θ }{sinθ - cosθ} \\ \frac{(cosθ + sinθ)( {sin}^{2}θ + {cos}^{2} θ - sin θ cosθ )}{(cosθ + sinθ)} + \frac{(sinθ - cosθ)( {sin}^{2}θ + {cos}^{2} θ + sinθ cos θ ) }{(sinθ - cosθ ) } \\ = 1 - sin θ cosθ + 1 + sin θ cosθ \\ = 2$