Prove that : $\displaystyle 2\left(\sin ^{6} \theta+\cos ^{6} \theta\right)-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)+1=0$.

\[\begin{aligned} \text { LHS } & =2\left(\sin ^{6} \theta+\cos ^{6} \theta\right)-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)+1 \\ & =2\left[\left(\sin ^{2} \theta\right)^{3}+\left(\cos ^{2} \theta\right)^{3}\right]-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)+1 \\ & =2\left[\left(\sin ^{2} \theta+\cos ^{2} \theta\right)\left(\sin ^{4} \theta-\sin ^{2} \theta \cos ^{2} \theta+\cos ^{4} \theta\right)\right]- \\ & \quad 3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)+1 \end{aligned} \] \[\begin{aligned} & =2\left[\sin ^{4} \theta+\cos ^{4} \theta-\sin ^{2} \theta \cos ^{2} \theta\right]-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)+1 \\ & =-\left[\sin ^{4} \theta+\cos ^{4} \theta+2 \sin ^{2} \theta \cos ^{2} \theta\right]+1 \\ & =-\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2}+1 \\ & =-1+1=0 \end{aligned} \]