The value of the expression 2(sin^6 theta + c | vedclass

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(B) Let $E = 2(\sin^6 	heta + \cos^6 	heta) - 3(\sin^4 	heta + \cos^4 	heta) + 1$.We know that $\sin^2 	heta + \cos^2 	heta = 1$.First,express $

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$0$

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(B) Let $E = 2(\sin^6 	heta + \cos^6 	heta) - 3(\sin^4 	heta + \cos^4 	heta) + 1$.We know that $\sin^2 	heta + \cos^2 	heta = 1$.First,express $\sin^4 	heta + \cos^4 	heta$ as:$\sin^4 	heta + \cos^4 	heta = (\sin^2 	heta + \cos^2 	heta)^2 - 2\sin^2 	heta \cos^2 	heta = 1 - 2\sin^2 	heta \cos^2 	heta$.Next,express $\sin^6 	heta + \cos^6 	heta$ as:$\sin^6 	heta + \cos^6 	heta = (\sin^2 	heta)^3 + (\cos^2 	heta)^3 = (\sin^2 	heta + \cos^2 	heta)(\sin^4 	heta + \cos^4 	heta - \sin^2 	heta \cos^2 	heta)$.Substituting $\sin^4 	heta + \cos^4 	heta = 1 - 2\sin^2 	heta \cos^2 	heta$,we get:$\sin^6 	heta + \cos^6 	heta = 1 \cdot (1 - 2\sin^2 	heta \cos^2 	heta - \sin^2 	heta \cos^2 	heta) = 1 - 3\sin^2 	heta \cos^2 	heta$.Now,substitute these into the expression $E$:$E = 2(1 - 3\sin^2 	heta \cos^2 	heta) - 3(1 - 2\sin^2 	heta \cos^2 	heta) + 1$.$E = 2 - 6\sin^2 	heta \cos^2 	heta - 3 + 6\sin^2 	heta \cos^2 	heta + 1$.$E = 2 - 3 + 1 = 0$.Thus,the value of the expression is $0$.

The value of the expression $2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta) + 1$ is

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