If $\sin \theta+\sin ^{2} \theta+\sin ^{3} \theta=1$, then $\cos ^{6} \theta-4 \cos ^{4} \theta+8$

$\cos ^{2} \theta+4=?$

$\sin \theta+\sin ^{2} \theta+\sin ^{3} \theta=1$

$\Rightarrow \sin \theta+\sin ^{3} \theta=1-\sin ^{2} \theta$

$\Rightarrow \sin \theta\left(1+\sin ^{2} \theta\right)=\cos ^{2} \theta$

$\Rightarrow \sin ^{2} \theta\left(1+\sin ^{2} \theta\right)^{2}=\cos ^{4} \theta$

$\Rightarrow\left(1-\cos ^{2} \theta\right)\left\{1+\left(1-\cos ^{2} \theta\right)\right\}^{2}=\cos ^{4} \theta$

$\Rightarrow\left(1-\cos ^{2} \theta\right)\left(2-\cos ^{2} \theta\right)^{2}=\cos ^{4} \theta$

$\Rightarrow\left(1-\cos ^{2} \theta\right)\left(4-4 \cos ^{2} \theta+\cos ^{4} \theta=\cos ^{4} \theta\right)$

$\Rightarrow 4-4 \cos ^{2} \theta+\cos ^{4} \theta-4 \cos ^{2} \theta+4 \cos ^{4} \theta-\cos ^{6} \theta=\cos ^{4} \theta$

$\Rightarrow-\cos ^{6} \theta+4 \cos ^{4} \theta-8 \cos ^{2} \theta+4=0$

$\Rightarrow \cos ^{6} \theta-4 \cos ^{4} \theta+8 \cos ^{2} \theta=4$

$\Rightarrow \cos ^{6} \theta-4 \cos ^{4} \theta+8 \cos ^{2} \theta+4=8$