Menu Close

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-

Tinku Tara 0 Comments
Pin




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

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *