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Prove: $\sin ^{6} \theta+\cos ^{6} \theta=1-3 \sin ^{2} \theta \cos ^{2} \theta$.
Given: $\sin ^{6} \theta+\cos ^{6} \theta=1-3 \sin ^{2} \theta \cos ^{2} \theta$.
To do: To prove that $L.H.S.=R.H.S.$
Solution:
$L.H.S.=sin^6\theta+cos^6\theta$
$=(sin^2\theta)^3+(cos^2\theta)^3$
Let $sin^2\theta=a$ and $cos^2\theta=b$
$\therefore L.H.S.=a^3+b^3$
$=( a+b)^3-3ab( a+b)$
$=(sin^2\theta-cos^2\theta)^3-3sin^2\theta cos^2\theta(sin^2\theta+cos^2\theta)$
$=( 1)^3-3sin^2\theta cos^2\theta$ $[\because sin^2\theta+cos^2\theta=1]$
$=1-3sin^2\theta cos^2\theta$
$=R.H.S.$
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