Correct Option: B

2 (sin⁶θ + cos⁶θ) – 3 (sin⁴θ + cos⁴θ) + 1
= 2[(sin²θ)3 + (cos²θ)3] – 3 [(sin²θ)2 + (cos²θ)2] + 1
= 2 (sin²θ + cos²θ) (sin⁴θ + cos⁴θ – sin²θ . cos²θ) – 3 [(sin²θ + cos²θ)² – 2sin²θ . cos²θ] + 1
[∵ a³ + b³ = (a + b) (a² – ab + b²) ; a² + b² = (a + b)² – 2ab]
= 2 sin4θ + 2 cos4θ – 2sin²θ . cos²θ – 3 + 6 sin²θ . cos²θ + 1
= 2 [(sin²θ + cos²θ)2 – 2 sin²θ . cos²θ] – 3 + 4 sin²θ . cos²θ + 1
= 2 – 4 sin²θ . cos²θ – 3 + 4 sin²θ cos²θ + 1 = 2 – 3 + 1
= 0