sin ^{6} theta+cos ^{6} theta is equal to | vedclass

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

Vedclass · Accepted Answer

$1-3 \sin ^{2} 	heta \cos ^{2} 	heta$

Vedclass · Answer

(C) We know that $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$.Let $a = \sin^2 	heta$ and $b = \cos^2 	heta$.Then,$\sin^6 	heta + \cos^6 	heta = (\sin^2 	heta)^3 + (\cos^2 	heta)^3$.Using the identity,we get:$\sin^6 	heta + \cos^6 	heta = (\sin^2 	heta + \cos^2 	heta)^3 - 3(\sin^2 	heta)(\cos^2 	heta)(\sin^2 	heta + \cos^2 	heta)$.Since $\sin^2 	heta + \cos^2 	heta = 1$,we substitute this value:$\sin^6 	heta + \cos^6 	heta = (1)^3 - 3 \sin^2 	heta \cos^2 	heta (1)$.Therefore,$\sin^6 	heta + \cos^6 	heta = 1 - 3 \sin^2 	heta \cos^2 	heta$.

$\sin ^{6} \theta+\cos ^{6} \theta$ is equal to

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