Explanation

For identity 1, factor the numerator: sin³θ + cos³θ = (sinθ + cosθ)(sin²θ − sinθcosθ + cos²θ) = (sinθ + cosθ)(1 − sinθcosθ). Dividing by (sinθ + cosθ) leaves 1 − sinθcosθ, and adding sinθcosθ gives 1 — true. For identity 2, 1 − sin⁶θ = (1 − sin²θ)(1 + sin²θ + sin⁴θ) = cos²θ(1 + sin²θ + sin⁴θ); since 1 + sin²θ + sin⁴θ = cos⁴θ + 3sin²θ (substitute cos²θ = 1 − sin²θ to verify), identity 2 is also true. Both hold, so the answer is (C).