Prove that 2(sin ⁶θ + cos ⁶θ ) - 3(sin ⁴θ +cos ⁴θ ) + 1 = 0.

2(sin ⁶ θ +cos ⁶ θ ) – 3(sin ⁴ θ + cos ⁴ θ ) + 1

= 2[(sin ² θ ) ³ + (cos ² θ ) ³ ] – 3[(sin ² θ ) ² + (cos ² θ ) ² ] + 1
= 2[(sin ² θ + cos ² θ ) ³ – 3sin ² θ cos ² θ (sin ² θ + cos ² θ )] – 3[(sin ² θ + cos ² θ ) ² – 2sin ² θ cos ² θ ] + 1
[The algebraic identity a ³ + b ³ = (a + b) ³ – 3ab(a + b) and a ² + b ² = (a + b) ² – 2ab
are used in the above step where a = sin ² θ and b = cos ² θ].

Since, sin ² θ + cos ² θ = 1, we have
L.H.S. = 2[ 1 – 3 sin ² θ cos ² θ ] – 3[1– 2 sin ² θ cos ² θ ] + 1
= 2 – 6 sin ² θ cos ² θ - 3 + 6 sin ² θ cos ² θ + 1
= - 3 + 3 = 0