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Question-223110

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Question Number 223110 by Rojarani last updated on 15/Jul/25
Answered by fantastic last updated on 15/Jul/25
(((x^(11) −(1/x^(11) ))−(x^9 −(1/x^9 )))/((x^(10) −(1/x^(10) ))−(x^8 −(1/x^8 )))) First the upper portion x^(11) −(1/x^(11) )−x^9 +(1/x^9 ) =x^9 (x^2 −1)+(1/x^(11) )(−1+x^2 ) =(x^2 −1)(x^9 +(1/x^(11) )) Now the lower portion x^(10) −(1/x^(10) )−x^8 +(1/x^8 ) =x^8 (x^2 −1)+(1/x^(10) )(x^2 −1) =(x^2 −1)(x^8 +(1/x^(10) )) those two (x^2 −1) will cancel out So the fraction becomes (((x^9 +(1/x^(11) )))/((x^8 +(1/x^(10) )))) =((x^4 (x^5 +(1/x^(15) )))/(x^4 (x^4 +(1/x^(14) )))) =(((x^5 +(1/x^(15) )))/((x^4 +(1/x^(14) )))) =(((x^5 +(1/x^(15) )))/(x(x^3 +(1/x^(15) ))))=((((x^(20) +1))/x^(15) )/((x(x^(18) +1))/x^(15) ))=((x^(20) +1)/(x(x^(18) +1))) Now x^5 +(1/x^5 )=123 or x^(10) +1=123x^5 or x^(20) +1+2x^(10) =123^2 x^(10) ∴x^(20) +1=x^(10) (123^2 −2) So the fraction becomes ((x^(10) (123^2 −2))/(x(x^(18) +1)))=((x^9 (123^2 −2))/((x^(18) +1))) =((x^9 (123^2 −2)×(1/x^9 ))/((x^(18) +1)×(1/x^9 ))) =((123^2 −2)/(x^9 +(1/x^9 ))) =((15127)/(x^9 +(1/x^9 ))) Now by using binomial theorem we will get x^5 +(1/x^5 )=(x+(1/x))^5 −5(x+(1/x))^3 +5(x+(1/x)) let x+(1/x) be a so the equation becomes a^5 −5a^3 +5a=123 a=3 satisfies the equation so( x+(1/x)) =3 Now x^3 +(1/x^3 )=(x+(1/x))^3 −3(x+(1/x))⇒27−9=18 So x^9 +(1/x^9 )=(x^3 )^3 +((1/x^3 ))^3 =(x^3 +(1/x^3 ))^3 −3(x^3 +(1/x^3 ))⇒(18)^3 −3×18=5778 So ((15127)/(x^9 +(1/x^9 )))=((15127)/(5778)) determinant (((So (((x^(11) −(1/x^(11) ))−(x^9 −(1/x^9 )))/((x^(10) −(1/x^(10) ))−(x^8 −(1/x^8 ))))=((15127)/(5778)))))✓
$$\frac{\left({x}^{\mathrm{11}} −\frac{\mathrm{1}}{{x}^{\mathrm{11}} }\right)−\left({x}^{\mathrm{9}} −\frac{\mathrm{1}}{{x}^{\mathrm{9}} }\right)}{\left({x}^{\mathrm{10}} −\frac{\mathrm{1}}{{x}^{\mathrm{10}} }\right)−\left({x}^{\mathrm{8}} −\frac{\mathrm{1}}{{x}^{\mathrm{8}} }\right)} \\ $$$${First}\:{the}\:{upper}\:{portion} \\ $$$${x}^{\mathrm{11}} −\frac{\mathrm{1}}{{x}^{\mathrm{11}} }−{x}^{\mathrm{9}} +\frac{\mathrm{1}}{{x}^{\mathrm{9}} } \\ $$$$={x}^{\mathrm{9}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)+\frac{\mathrm{1}}{{x}^{\mathrm{11}} }\left(−\mathrm{1}+{x}^{\mathrm{2}} \right) \\ $$$$=\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}^{\mathrm{9}} +\frac{\mathrm{1}}{{x}^{\mathrm{11}} }\right) \\ $$$${Now}\:{the}\:{lower}\:{portion} \\ $$$${x}^{\mathrm{10}} −\frac{\mathrm{1}}{{x}^{\mathrm{10}} }−{x}^{\mathrm{8}} +\frac{\mathrm{1}}{{x}^{\mathrm{8}} } \\ $$$$={x}^{\mathrm{8}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)+\frac{\mathrm{1}}{{x}^{\mathrm{10}} }\left({x}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$=\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}^{\mathrm{8}} +\frac{\mathrm{1}}{{x}^{\mathrm{10}} }\right) \\ $$$${those}\:{two}\:\left({x}^{\mathrm{2}} −\mathrm{1}\right)\:{will}\:{cancel}\:{out}\:{So}\:{the}\:{fraction}\:{becomes} \\ $$$$\frac{\left({x}^{\mathrm{9}} +\frac{\mathrm{1}}{{x}^{\mathrm{11}} }\right)}{\left({x}^{\mathrm{8}} +\frac{\mathrm{1}}{{x}^{\mathrm{10}} }\right)} \\ $$$$=\frac{\cancel{{x}^{\mathrm{4}} }\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{15}} }\right)}{\cancel{{x}^{\mathrm{4}} }\left({x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{14}} }\right)} \\ $$$$=\frac{\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{15}} }\right)}{\left({x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{14}} }\right)} \\ $$$$=\frac{\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{15}} }\right)}{{x}\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{15}} }\right)}=\frac{\frac{\left({x}^{\mathrm{20}} +\mathrm{1}\right)}{\cancel{{x}^{\mathrm{15}} }}}{\frac{{x}\left({x}^{\mathrm{18}} +\mathrm{1}\right)}{\cancel{{x}^{\mathrm{15}} }}}=\frac{{x}^{\mathrm{20}} +\mathrm{1}}{{x}\left({x}^{\mathrm{18}} +\mathrm{1}\right)} \\ $$$${Now}\:{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=\mathrm{123} \\ $$$${or}\:{x}^{\mathrm{10}} +\mathrm{1}=\mathrm{123}{x}^{\mathrm{5}} \\ $$$${or}\:{x}^{\mathrm{20}} +\mathrm{1}+\mathrm{2}{x}^{\mathrm{10}} =\mathrm{123}^{\mathrm{2}} {x}^{\mathrm{10}} \\ $$$$\therefore{x}^{\mathrm{20}} +\mathrm{1}={x}^{\mathrm{10}} \left(\mathrm{123}^{\mathrm{2}} −\mathrm{2}\right) \\ $$$${So}\:{the}\:{fraction}\:{becomes} \\ $$$$\frac{{x}^{\mathrm{10}} \left(\mathrm{123}^{\mathrm{2}} −\mathrm{2}\right)}{{x}\left({x}^{\mathrm{18}} +\mathrm{1}\right)}=\frac{{x}^{\mathrm{9}} \left(\mathrm{123}^{\mathrm{2}} −\mathrm{2}\right)}{\left({x}^{\mathrm{18}} +\mathrm{1}\right)} \\ $$$$=\frac{{x}^{\mathrm{9}} \left(\mathrm{123}^{\mathrm{2}} −\mathrm{2}\right)×\frac{\mathrm{1}}{{x}^{\mathrm{9}} }}{\left({x}^{\mathrm{18}} +\mathrm{1}\right)×\frac{\mathrm{1}}{{x}^{\mathrm{9}} }} \\ $$$$=\frac{\mathrm{123}^{\mathrm{2}} −\mathrm{2}}{{x}^{\mathrm{9}} +\frac{\mathrm{1}}{{x}^{\mathrm{9}} }} \\ $$$$=\frac{\mathrm{15127}}{{x}^{\mathrm{9}} +\frac{\mathrm{1}}{{x}^{\mathrm{9}} }} \\ $$$${Now}\:{by}\:{using}\:{binomial}\:{theorem}\:{we}\:{will}\:{get} \\ $$$${x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{5}} −\mathrm{5}\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} +\mathrm{5}\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$${let}\:{x}+\frac{\mathrm{1}}{{x}}\:{be}\:{a}\:{so}\:{the}\:{equation}\:{becomes} \\ $$$${a}^{\mathrm{5}} −\mathrm{5}{a}^{\mathrm{3}} +\mathrm{5}{a}=\mathrm{123} \\ $$$$\:{a}=\mathrm{3}\:\:{satisfies}\:{the}\:{equation} \\ $$$${so}\left(\:{x}+\frac{\mathrm{1}}{{x}}\right)\:=\mathrm{3} \\ $$$${Now}\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} −\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right)\Rightarrow\mathrm{27}−\mathrm{9}=\mathrm{18} \\ $$$${So}\:{x}^{\mathrm{9}} +\frac{\mathrm{1}}{{x}^{\mathrm{9}} }=\left({x}^{\mathrm{3}} \right)^{\mathrm{3}} +\left(\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\mathrm{3}} =\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\mathrm{3}} −\mathrm{3}\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)\Rightarrow\left(\mathrm{18}\right)^{\mathrm{3}} −\mathrm{3}×\mathrm{18}=\mathrm{5778} \\ $$$${So}\:\frac{\mathrm{15127}}{{x}^{\mathrm{9}} +\frac{\mathrm{1}}{{x}^{\mathrm{9}} }}=\frac{\mathrm{15127}}{\mathrm{5778}} \\ $$$$\begin{array}{|c|}{{So}\:\frac{\left({x}^{\mathrm{11}} −\frac{\mathrm{1}}{{x}^{\mathrm{11}} }\right)−\left({x}^{\mathrm{9}} −\frac{\mathrm{1}}{{x}^{\mathrm{9}} }\right)}{\left({x}^{\mathrm{10}} −\frac{\mathrm{1}}{{x}^{\mathrm{10}} }\right)−\left({x}^{\mathrm{8}} −\frac{\mathrm{1}}{{x}^{\mathrm{8}} }\right)}=\frac{\mathrm{15127}}{\mathrm{5778}}}\\\hline\end{array}\checkmark \\ $$

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