Prove the following by using trigonometric identities:

2(sin⁶θ + cos⁶θ) — 3(sin⁴θ + cos⁴θ) + 1 = 0

2((sin²θ)³ + (cos²θ)³) — 3(sin⁴θ + cos⁴θ) + 1

(∵ a² + b² = (a + b)²–2ab)

⇒ 2[(sin²θ + cos²θ)(sin⁴θ –sin²θcos²θ + cos⁴θ)]–3[(sin²θ + cos²θ)–2sin²θcos²θ] + 1

= 2[(sin⁴θ –sin²θcos²θ + cos⁴θ)]–3[1–2sin²θcos²θ] + 1

= 2sin⁴θ–2sin²θcos²θ + 2cos⁴θ–3 + 6sin²θcos²θ + 1

= 2(sin⁴θ + 4sin²θcos²θ + cos⁴θ)–2

= 2(sin²θ + cos²θ)²–2

= 2–2

= 0