2(sin^6 theta + cos^6 theta) - 3(sin^4 theta | vedclass

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(A) We know that $\sin^2 	heta + \cos^2 	heta = 1$. Using the identity $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$,we have: $\sin^6 	heta + \cos^6 	heta = (\

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$-1$

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(A) We know that $\sin^2 	heta + \cos^2 	heta = 1$. Using the identity $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$,we have: $\sin^6 	heta + \cos^6 	heta = (\sin^2 	heta)^3 + (\cos^2 	heta)^3 = (\sin^2 	heta + \cos^2 	heta)^3 - 3 \sin^2 	heta \cos^2 	heta (\sin^2 	heta + \cos^2 	heta) = 1 - 3 \sin^2 	heta \cos^2 	heta$. Using the identity $a^2 + b^2 = (a+b)^2 - 2ab$,we have: $\sin^4 	heta + \cos^4 	heta = (\sin^2 	heta)^2 + (\cos^2 	heta)^2 = (\sin^2 	heta + \cos^2 	heta)^2 - 2 \sin^2 	heta \cos^2 	heta = 1 - 2 \sin^2 	heta \cos^2 	heta$. Substituting these into the expression: $2(1 - 3 \sin^2 	heta \cos^2 	heta) - 3(1 - 2 \sin^2 	heta \cos^2 	heta)$ $= 2 - 6 \sin^2 	heta \cos^2 	heta - 3 + 6 \sin^2 	heta \cos^2 	heta$ $= -1$.

$2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta) = $

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