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If $\sin ^{6} \theta+\cos ^{6} \theta=\frac{1}{4}$
then $\frac{1}{\sin ^{6} \theta}+\frac{1}{\cos ^{6} \theta}=?$
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$\sin ^{6} \theta+\cos ^{6} \theta=\frac{1}{4} ; \frac{1}{\sin ^{6} \theta}+\frac{1}{\cos ^{6} \theta}=$ ?

We have $\sin ^{6} \theta+\cos ^{6} \theta=\frac{1}{4}$.
\begin{aligned} &\Rightarrow\left(\sin ^{2} \theta\right)^{3}+\left(\cos ^{2} \theta\right)^{3}=\frac{1}{4} \\ &\Rightarrow\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2}-3 \sin ^{2} \theta \cos ^{2} \theta\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=\frac{1}{4} \\ &\Rightarrow 1-3 \sin ^{2} \theta \cos ^{2} \theta=\frac{1}{4} \Rightarrow 3 \sin ^{2} \theta \cos ^{2} \theta=\frac{3}{4} \\ &\therefore \sin ^{2} \theta \cos ^{2} \theta=\frac{1}{4} \Rightarrow \sin ^{6} \theta \cos ^{6} \theta=\frac{1}{64} . \\ &\text { Thus, } \frac{1}{\sin ^{6} \theta}+\frac{1}{\cos ^{6} \theta}=\frac{\sin ^{6} \theta+\cos ^{6} \theta}{\sin ^{6} \theta \cos ^{6} \theta}=\frac{1}{\frac{1}{1}}=16 . \end{aligned}
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