The value of sin ^6(theta) + cos ^6(theta) + | vedclass

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

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(B) We use the algebraic identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$.
Let $a = \sin^2 	heta$ and $b = \cos^2 	heta$.
Then $\sin^6 	heta + \cos^6 	heta = (\sin^2 	heta)^3 + (\cos^2 	heta)^3$.
$= (\sin^2 	heta + \cos^2 	heta)((\sin^2 	heta)^2 - \sin^2 	heta \cos^2 	heta + (\cos^2 	heta)^2)$.
Since $\sin^2 	heta + \cos^2 	heta = 1$,we have:
$= 1 \cdot (\sin^4 	heta - \sin^2 	heta \cos^2 	heta + \cos^4 	heta)$.
We know that $\sin^4 	heta + \cos^4 	heta = (\sin^2 	heta + \cos^2 	heta)^2 - 2 \sin^2 	heta \cos^2 	heta = 1 - 2 \sin^2 	heta \cos^2 	heta$.
Substituting this back:
$\sin^6 	heta + \cos^6 	heta = 1 - 2 \sin^2 	heta \cos^2 	heta - \sin^2 	heta \cos^2 	heta = 1 - 3 \sin^2 	heta \cos^2 	heta$.
Therefore,$\sin^6 	heta + \cos^6 	heta + 3 \sin^2 	heta \cos^2 	heta = (1 - 3 \sin^2 	heta \cos^2 	heta) + 3 \sin^2 	heta \cos^2 	heta = 1$.

The value of $\sin ^6(\theta) + \cos ^6(\theta) + 3 \sin ^2(\theta) \cos ^2(\theta)$ is

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