Prove that 2(sin6θ+cos6θ)-3(sin4θ +cos4θ)+1=0

Question : Prove that 2(sin⁶θ+cos⁶θ)-3(sin⁴θ +cos⁴θ)+1=0

Doubt by Muskan 

Solution :
2(sin⁶θ+cos⁶θ)-3(sin⁴θ +cos⁴θ)+1=0

LHS
=2(sin⁶θ+cos⁶θ)-3(sin⁴θ +cos⁴θ)+1
=2[(sin²θ)³+(cos²θ)³] - 3(sin⁴θ +cos⁴θ)+1
=2[(sin²θ+cos²θ)(sin⁴θ+cos⁴θ-sin²θcos²θ)]-3(sin⁴θ +cos⁴θ)+1
[∵ (a³+b³)=(a+b)(a²+b²-ab)]
=2[(1)(sin⁴θ+cos⁴θ-sin²θcos²θ)]-3(sin⁴θ +cos⁴θ)+1
[∵ sin²θ+cos²θ = 1]

=2sin⁴θ+2cos⁴θ-2sin²θcos²θ-3sin⁴θ-3cos⁴θ+1
=-sin⁴θ-cos⁴θ-2sin²θcos²θ+1
=-(sin⁴θ+cos⁴θ+2sin²θcos²θ)+1
=-(sin²θ+cos²θ)2+1
[∵ sin²θ+cos²θ = 1]
=-(1)2+1
=-1+1
=0 
= RHS

∴ LHS = RHS
Hence Proved.