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Question Number 176598 by Rasheed.Sindhi last updated on 22/Sep/22

x^3 +(1/x^3 )=1  (((x^5 +(1/x^5 ))^3 −1)/(x^5 +(1/x^5 )))=?  Q#176387 reposted for a new answer.

$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\mathrm{1} \\ $$$$\frac{\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }}=? \\ $$$${Q}#\mathrm{176387}\:{reposted}\:{for}\:{a}\:{new}\:{answer}. \\ $$

Answered by Rasheed.Sindhi last updated on 22/Sep/22

 determinant (((x^3 +(1/x^3 )=1)))  (x^3 +(1/x^3 ))^3 =(1)^3   x^9 +(1/x^9 )+3(x^3 +(1/x^3 ))=1  x^9 +(1/x^9 )+3(1)=1  x^9 +(1/x^9 )=1−3=−2   determinant (((x^9 +(1/x^9 )=−2)))  (x^3 +(1/x^3 ))^5 =x^(15) +(1/x^(15) )+5(x^9 +(1/x^9 ))+10(x^3 +(1/x^3 ))  (1)^5 =x^(15) +(1/x^(15) )+5(−2)+10(1)  x^(15) +(1/x^(15) )=1                    determinant (((x^(15) +(1/x^(15) )=1)))  (x^5 +(1/x^5 ))^3 =x^(15) +(1/x^(15) )+3(x^5 +(1/x^5 ))    Let x^5 +(1/x^5 )=u  u^3 =3u+1    (((x^5 +(1/x^5 ))^3 −1)/(x^5 +(1/x^5 )))=((u^3 −1)/u)=((3u+1−1)/u)=3

$$\begin{array}{|c|}{{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\mathrm{1}}\\\hline\end{array} \\ $$$$\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\mathrm{3}} =\left(\mathrm{1}\right)^{\mathrm{3}} \\ $$$${x}^{\mathrm{9}} +\frac{\mathrm{1}}{{x}^{\mathrm{9}} }+\mathrm{3}\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)=\mathrm{1} \\ $$$${x}^{\mathrm{9}} +\frac{\mathrm{1}}{{x}^{\mathrm{9}} }+\mathrm{3}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$${x}^{\mathrm{9}} +\frac{\mathrm{1}}{{x}^{\mathrm{9}} }=\mathrm{1}−\mathrm{3}=−\mathrm{2}\:\:\begin{array}{|c|}{{x}^{\mathrm{9}} +\frac{\mathrm{1}}{{x}^{\mathrm{9}} }=−\mathrm{2}}\\\hline\end{array} \\ $$$$\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\mathrm{5}} ={x}^{\mathrm{15}} +\frac{\mathrm{1}}{{x}^{\mathrm{15}} }+\mathrm{5}\left({x}^{\mathrm{9}} +\frac{\mathrm{1}}{{x}^{\mathrm{9}} }\right)+\mathrm{10}\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right) \\ $$$$\left(\mathrm{1}\right)^{\mathrm{5}} ={x}^{\mathrm{15}} +\frac{\mathrm{1}}{{x}^{\mathrm{15}} }+\mathrm{5}\left(−\mathrm{2}\right)+\mathrm{10}\left(\mathrm{1}\right) \\ $$$${x}^{\mathrm{15}} +\frac{\mathrm{1}}{{x}^{\mathrm{15}} }=\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\begin{array}{|c|}{{x}^{\mathrm{15}} +\frac{\mathrm{1}}{{x}^{\mathrm{15}} }=\mathrm{1}}\\\hline\end{array} \\ $$$$\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{3}} ={x}^{\mathrm{15}} +\frac{\mathrm{1}}{{x}^{\mathrm{15}} }+\mathrm{3}\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right) \\ $$$$\:\:{Let}\:{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }={u} \\ $$$${u}^{\mathrm{3}} =\mathrm{3}{u}+\mathrm{1} \\ $$$$ \\ $$$$\frac{\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }}=\frac{{u}^{\mathrm{3}} −\mathrm{1}}{{u}}=\frac{\mathrm{3}{u}+\mathrm{1}−\mathrm{1}}{{u}}=\mathrm{3} \\ $$

Commented by Tawa11 last updated on 23/Sep/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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