Prove that $\displaystyle \sin ^{6} \theta+\cos ^{6} \theta=1-3 \sin ^{2} \theta \cos ^{2} \theta$.

\[\begin{aligned} & \text { LHS }=\sin ^{6} \theta+\cos ^{6} \theta \\ & =\left(\sin ^{2} \theta\right)^{3}+\left(\cos ^{2} \theta\right)^{3} \\ & =\left(\sin ^{2} \theta+\cos ^{2} \theta\right)\left[\left(\sin ^{2} \theta\right)^{2}+\left(\cos ^{2} \theta\right)^{2}-\sin ^{2} \theta \cos ^{2} \theta\right] \\ & =\sin ^{4} \theta+\cos ^{4} \theta-\sin ^{2} \theta \cos ^{2} \theta \\ & =\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2}-2 \sin ^{2} \theta \cos ^{2} \theta-\sin ^{2} \theta \cos ^{2} \theta \\ & =1-3 \sin ^{2} \theta \cos ^{2} \theta \\ & =\text { RHS } \end{aligned} \]