1. **State the problem:** Prove the trigonometric identity $$(\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta) \equiv \sin^3 \theta + \cos^3 \theta.$$\n\n2. **Recall the formula:** The right side is a sum of cubes, which can be factored as $$a^3 + b^3 = (a + b)(a^2 - ab + b^2).$$ Here, $a = \sin \theta$ and $b = \cos \theta$.\n\n3. **Apply the sum of cubes factorization:**\n$$\sin^3 \theta + \cos^3 \theta = (\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta).$$\n\n4. **Use the Pythagorean identity:**\nSince $$\sin^2 \theta + \cos^2 \theta = 1,$$ substitute this into the factorization:\n$$\sin^3 \theta + \cos^3 \theta = (\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta).$$\n\n5. **Compare both sides:**\nThe left side is exactly $$(\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta),$$ which matches the right side after factorization.\n\n6. **Conclusion:**\nTherefore, the identity is proven: $$(\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta) \equiv \sin^3 \theta + \cos^3 \theta.$$