sin^6 theta + cos^6 theta = 1-3 sin^2 theta cos^2 theta?

sin⁶θ + cos⁶θ = (sin²θ)³ + (cos²θ)³
= 1(sin²θ + cos²θ) (sin⁴θ + sin²θcos²θ + cos⁴θ)
= (sin⁴θ + sin²θcos²θ + cos⁴θ)
= (sin²θ(sin²θ + cos²θ) + cos⁴θ)
= sin²θ·1 + cos⁴θ
= (1-cos²θ) + cos⁴θ
= 1 - 3sin²θcos²θ

cos²(θ) + sin²(θ) = 1

[cos²(θ) + sin²(θ)]³ = [1]³

[cos²(θ) + sin²(θ)]³ = 1

[cos²(θ) + sin²(θ)]².[cos²(θ) + sin²(θ)] = 1

[cos⁴(θ) + 2.cos²(θ).sin²(θ) + sin⁴(θ)].[cos²(θ) + sin²(θ)] = 1

cos⁶(θ) + cos⁴(θ).sin²(θ) + 2.cos⁴(θ).sin²(θ) + 2.cos²(θ).sin⁴(θ) + cos²(θ).sin⁴(θ) + sin⁶(θ) = 1

cos⁶(θ) + 3.cos⁴(θ).sin²(θ) + 3.cos²(θ).sin⁴(θ) + sin⁶(θ) = 1

cos⁶(θ) + [3.cos⁴(θ).sin²(θ) + 3.cos²(θ).sin⁴(θ)] + sin⁶(θ) = 1

cos⁶(θ) + 3.cos²(θ).sin²(θ).[cos²(θ) + sin²(θ)] + sin⁶(θ) = 1 → recall: cos²(θ) + sin²(θ) = 1

cos⁶(θ) + 3.cos²(θ).sin²(θ) + sin⁶(θ) = 1

cos⁶(θ) + sin⁶(θ) = 1 - 3.cos²(θ).sin²(θ)

L.H.S.
= sin⁶θ + cos⁶θ
= (sin²θ)³ + (cos²θ)³
= (sin²θ + cos²θ)[(sin²θ)² - sin²θ cos²θ + (cos²θ)²] ...... for u³ + v³ = (u + v)(u² + uv + v²)
= (sin²θ + cos²θ)[sin⁴θ - sin²θ cos²θ + cos⁴θ]
= (1)[sin⁴θ - sin²θ cos²θ + cos⁴θ]
= (sin²θ)²+ (cos²θ)² - sin²θ cos²θ
= [(sin²θ)²+ 2 sin²θ cos²θ + (cos²θ)²] - 3 sin²θ cos²θ
= [sin²θ + cos²θ]² - 3 sin²θ cos²θ
= 1 - 3 sin²θ cos²θ
= R.H.S.

Hence, sin⁶θ + cos⁶θ = 1 - 3 sin²θ cos²θ