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

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Use \(\sin^2\theta+\cos^2\theta=1\). Put \(a=\sin^2\theta\), \(b=\cos^2\theta\), then use \(a^3+b^3=(a+b)^3-3ab(a+b)\).

Updated On: May 7, 2026

\(0\)
\(1\)
\(2\)
\(-1\)

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The Correct Option is B

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