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 $cos^2\theta=b$

$\therefore L.H.S.=a^3+b^3$

$=( a+b)^3-3ab( a+b)$

$=(sin^2\theta-cos^2\theta)^3-3sin^2\theta cos^2\theta(sin^2\theta+cos^2\theta)$

$=( 1)^3-3sin^2\theta cos^2\theta$         $[\because sin^2\theta+cos^2\theta=1]$

$=1-3sin^2\theta cos^2\theta$

$=R.H.S.$