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Question Number 151959 by john_santu last updated on 24/Aug/21

Answered by iloveisrael last updated on 24/Aug/21

$${\left(\sin {}^{2}x\right)}^3+{\left(\cos {}^{2}x\right)}^3=\frac{1}{4}$$$$\Rightarrow \left(1\right)(\sin {}^{4}x+\cos {}^{4}x-{\left(\sin x\cos x\right)}^2)=\frac{1}{4}$$$$\Rightarrow {(\sin {}^{2}x+\cos {}^{2}x)}^2-3{\left(\sin x\cos x\right)}^2=\frac{1}{4}$$$$\Rightarrow {\left(\sin x\cos x\right)}^2=\frac{1-\frac{1}{4}}{3}=\frac{1}{4}$$$$\iff \frac{1}{\sin {}^{6}x}+\frac{1}{\cos {}^{6}x}=\frac{\frac{1}{4}}{{\left(\sin {}^{2}x\cos {}^{2}x\right)}^3}$$$$=\frac{1}{{\left(\frac{1}{4}\right)}^2}=16.$$

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