2 (sin^6 theta + cos ^6)-3(sin^4 theta + cos ^4 theta) is independent of theta
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2(sin⁶θ+cos⁶θ)-3(sin⁴θ+cos⁴θ)+1
= 2{(sin²θ)³+(cos²θ)³}-3{(sin²θ)²+(cos²θ)²}+1
= 2{(sin²θ+cos²θ)³-3sin²θcos²θ(sin²θ+cos²θ)}-3{(sin²θ+cos²θ)²-2sin²θcos²θ}+1
=2(1-3sin²θcos²θ)-3(1-2sin²θcos²θ)+1
=2-6sin²θcos²θ-3+6sin²θcos²θ+1
=-1+1
=0 (Proved)
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= 2{(sin²θ)³+(cos²θ)³}-3{(sin²θ)²+(cos²θ)²}+1
= 2{(sin²θ+cos²θ)³-3sin²θcos²θ(sin²θ+cos²θ)}-3{(sin²θ+cos²θ)²-2sin²θcos²θ}+1
=2(1-3sin²θcos²θ)-3(1-2sin²θcos²θ)+1
=2-6sin²θcos²θ-3+6sin²θcos²θ+1
=-1+1
=0 (Proved)
hope this helps you
please follow me
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