Q) Prove that, 2(sin⁶ θ + cos⁶ θ) – 3 (sin⁴ θ + cos⁴ θ) +1 = 0

Ans: We know that, (a-b)³ = a³ – b³ – 3ab (a – b)

And (a-b)² = a² + b² – 2ab

LHS: 2(sin⁶ θ + cos⁶ θ) – 3 (sin⁴ θ + cos⁴ θ) +1

= 2[sin⁶ θ + (1 – sin²θ)³] – 3 [sin⁴ θ + (1- sin ² θ)²] +1

= 2 [sin⁶ θ + [1 – sin⁶ θ – 3 sin² θ (1-sin² θ)] – 3 [sin⁴ θ + 1 + sin⁴ θ – 2 sin² θ] + 1

= 2[1- 3 sin² θ + 3 sin⁴ θ] – 3 [2 sin⁴ θ + 1 – 2 sin² θ] + 1

= 2 – 6 sin² θ + 6 sin⁴ θ – 6 sin⁴ θ – 3 + 6 sin² θ + 1

= 0     =    RHS……… Hence Proved