Explanation

This tests the sum-of-cubes identity in trigonometry. Write sin⁶x + cos⁶x = (sin²x)³ + (cos²x)³ = (sin²x + cos²x)³ - 3sin²xcos²x(sin²x + cos²x) = 1 - 3sin²xcos²x. From sin x + cos x = c, squaring gives sin x cos x = (c² - 1)/2. Substituting: 1 - 3(c²-1)²/4 = (1 + 6c² - 3c⁴)/4. Option (a) with denominator 16 is the trap from mishandling the square of (c²-1)/2. Hence option (b).