Question

sin^ 6 theta - cos^ 6 theta​

Answer 1

sin6θ+cos6θ+3sin2θcos2θ

⇒LHS=(sin2θ)3+(cos2θ)3+3sin2θcos2θ

Using, [a3+b3=(a+b)3−3ab(a+b)]

⇒LHS=(sin2θ+cos2θ)3−3sin2θcos2θ(sin2θ+cos2θ)3+3sin2θcos2θ

⇒LHS=1−3sin2θcos2θ+3sin2θcos2θ=1=RHS

Answer 2

This is the answer

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