sin^6 theta + cos^6 theta + 3 sin^2 theta cos | vedclass

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(C) We know the algebraic identity $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$.Let $a = \sin^2 	heta$ and $b = \cos^2 	heta$.Then,$\sin^6 	heta + \cos^6 \

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$1$

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(C) We know the algebraic identity $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$.Let $a = \sin^2 	heta$ and $b = \cos^2 	heta$.Then,$\sin^6 	heta + \cos^6 	heta = (\sin^2 	heta)^3 + (\cos^2 	heta)^3 = (\sin^2 	heta + \cos^2 	heta)^3 - 3 \sin^2 	heta \cos^2 	heta (\sin^2 	heta + \cos^2 	heta)$.Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\sin^6 	heta + \cos^6 	heta = (1)^3 - 3 \sin^2 	heta \cos^2 	heta (1) = 1 - 3 \sin^2 	heta \cos^2 	heta$.Substituting this into the original expression:$\sin^6 	heta + \cos^6 	heta + 3 \sin^2 	heta \cos^2 	heta = (1 - 3 \sin^2 	heta \cos^2 	heta) + 3 \sin^2 	heta \cos^2 	heta = 1$.Alternatively,by putting $	heta = 0^\circ$,we get $\sin^6(0^\circ) + \cos^6(0^\circ) + 3 \sin^2(0^\circ) \cos^2(0^\circ) = 0 + 1 + 0 = 1$.

$\sin^6 \theta + \cos^6 \theta + 3 \sin^2 \theta \cos^2 \theta = $

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