if-x-1-x-2-5-then-find-the-value-of-x-x-6-1-x-8-1-

Question Number 159961 by abdullah_ff last updated on 23/Nov/21

$$\mathrm{if}\:{x}\:+\:\frac{\mathrm{1}}{{x}}\:=\:\mathrm{2}\sqrt{\mathrm{5}}\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\frac{{x}\left({x}^{\mathrm{6}} \:−\:\mathrm{1}\right)}{{x}^{\mathrm{8}} \:−\:\mathrm{1}} \\ $$

Answered by Rasheed.Sindhi last updated on 23/Nov/21

A Hard & Difficult Approach! x + (1/x) = 2(√5) ; ((x(x^6 − 1))/(x^8 − 1))=? x^2 −2(√5) x+1=0 x=((2(√5) ±(√(20−4)))/2)=(√5) ±2 ▶x−1_( +2) =(√5) +1_( +2) ,(√5) −3_( +2) .........(i) x+1=(√5) +3,(√5) −1.........(ii) (i)×(ii): x^2 −1=8+4(√5) , 8−4(√5) .......(iii) x^2 +1=10+4(√5) , 10−4(√5) .....(iv) (iii)×(iv): x^4 −1=160+72(√5) , 160−72(√5) ...(v) x^4 +1=162+72(√5) , 162−72(√5)....(vi) (v)×(vi): x^8 −1=51840±23184(√5) ▶(i):x−1=1+(√5) ,−3+(√5) ........(vii) (iv):x^2 +1=10±4(√5) x^2 +x+1=(10±4(√5) )+(±2+(√5) ) =12+5(√5) ,8−3(√5) ....(viii) (vii)×(viii): x^3 −1=37+17(√5) ,−39+17(√5) .....(ix) x^3 +1=39+17(√5) , −37+17(√5) ....(x) (ix)×(x): x^6 −1=2888±1292(√5) ▶((x(x^6 − 1))/(x^8 − 1))=((((√5) ±2)(2888±1292(√5) ))/(51840±23184(√5) )) =((±12236+5472(√5))/(51840±23184(√5) )).((51840∓23184(√5) )/(51840∓23184(√5) )) =((−10944(√5))/((51840)^2 −(23184^2 .5))) =((−10944(√5))/(−103680)) =((19(√5) )/(180))

$$\:\:\:\:\:\underline{\mathbb{A}\:\mathbb{H}\mathrm{ard}\:\&\:\mathbb{D}\mathrm{ifficult}\:\mathbb{A}\mathrm{pproach}!} \\ $$$${x}\:+\:\frac{\mathrm{1}}{{x}}\:=\:\mathrm{2}\sqrt{\mathrm{5}}\:;\:\frac{{x}\left({x}^{\mathrm{6}} \:−\:\mathrm{1}\right)}{{x}^{\mathrm{8}} \:−\:\mathrm{1}}=? \\ $$$${x}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{5}}\:{x}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{2}\sqrt{\mathrm{5}}\:\pm\sqrt{\mathrm{20}−\mathrm{4}}}{\mathrm{2}}=\sqrt{\mathrm{5}}\:\pm\mathrm{2} \\ $$$$\blacktriangleright{x}\underset{\:\:\:\underline{+\mathrm{2}}} {−\mathrm{1}}=\sqrt{\mathrm{5}}\:\underset{\:\:\underline{+\mathrm{2}}} {+\mathrm{1}},\sqrt{\mathrm{5}}\:\underset{\:\:\:\:\underline{+\mathrm{2}}} {−\mathrm{3}}………\left({i}\right) \\ $$$$\:\:\:\:{x}+\mathrm{1}=\sqrt{\mathrm{5}}\:+\mathrm{3},\sqrt{\mathrm{5}}\:−\mathrm{1}………\left({ii}\right) \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$$\:\:\:\:{x}^{\mathrm{2}} −\mathrm{1}=\mathrm{8}+\mathrm{4}\sqrt{\mathrm{5}}\:,\:\mathrm{8}−\mathrm{4}\sqrt{\mathrm{5}}\:…….\left({iii}\right) \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} +\mathrm{1}=\mathrm{10}+\mathrm{4}\sqrt{\mathrm{5}}\:,\:\mathrm{10}−\mathrm{4}\sqrt{\mathrm{5}}\:…..\left({iv}\right) \\ $$$$\left({iii}\right)×\left({iv}\right): \\ $$$$\:\:\:\:\:{x}^{\mathrm{4}} −\mathrm{1}=\mathrm{160}+\mathrm{72}\sqrt{\mathrm{5}}\:,\:\mathrm{160}−\mathrm{72}\sqrt{\mathrm{5}}\:…\left({v}\right) \\ $$$$\:\:\:\:\:{x}^{\mathrm{4}} +\mathrm{1}=\mathrm{162}+\mathrm{72}\sqrt{\mathrm{5}}\:,\:\mathrm{162}−\mathrm{72}\sqrt{\mathrm{5}}….\left({vi}\right) \\ $$$$\left({v}\right)×\left({vi}\right): \\ $$$$\:\:\:\:\:\:{x}^{\mathrm{8}} −\mathrm{1}=\mathrm{51840}\pm\mathrm{23184}\sqrt{\mathrm{5}}\: \\ $$$$\blacktriangleright\left({i}\right):{x}−\mathrm{1}=\mathrm{1}+\sqrt{\mathrm{5}}\:,−\mathrm{3}+\sqrt{\mathrm{5}}\:……..\left({vii}\right) \\ $$$$\:\:\:\:\:\left({iv}\right):{x}^{\mathrm{2}} +\mathrm{1}=\mathrm{10}\pm\mathrm{4}\sqrt{\mathrm{5}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} +{x}+\mathrm{1}=\left(\mathrm{10}\pm\mathrm{4}\sqrt{\mathrm{5}}\:\right)+\left(\pm\mathrm{2}+\sqrt{\mathrm{5}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{12}+\mathrm{5}\sqrt{\mathrm{5}}\:,\mathrm{8}−\mathrm{3}\sqrt{\mathrm{5}}\:….\left({viii}\right) \\ $$$$\left({vii}\right)×\left({viii}\right): \\ $$$${x}^{\mathrm{3}} −\mathrm{1}=\mathrm{37}+\mathrm{17}\sqrt{\mathrm{5}}\:,−\mathrm{39}+\mathrm{17}\sqrt{\mathrm{5}}\:…..\left({ix}\right) \\ $$$${x}^{\mathrm{3}} +\mathrm{1}=\mathrm{39}+\mathrm{17}\sqrt{\mathrm{5}}\:,\:−\mathrm{37}+\mathrm{17}\sqrt{\mathrm{5}}\:….\left({x}\right) \\ $$$$\left({ix}\right)×\left({x}\right): \\ $$$${x}^{\mathrm{6}} −\mathrm{1}=\mathrm{2888}\pm\mathrm{1292}\sqrt{\mathrm{5}}\: \\ $$$$\blacktriangleright\frac{{x}\left({x}^{\mathrm{6}} \:−\:\mathrm{1}\right)}{{x}^{\mathrm{8}} \:−\:\mathrm{1}}=\frac{\left(\sqrt{\mathrm{5}}\:\pm\mathrm{2}\right)\left(\mathrm{2888}\pm\mathrm{1292}\sqrt{\mathrm{5}}\:\right)}{\mathrm{51840}\pm\mathrm{23184}\sqrt{\mathrm{5}}\:} \\ $$$$=\frac{\pm\mathrm{12236}+\mathrm{5472}\sqrt{\mathrm{5}}}{\mathrm{51840}\pm\mathrm{23184}\sqrt{\mathrm{5}}\:}.\frac{\mathrm{51840}\mp\mathrm{23184}\sqrt{\mathrm{5}}\:}{\mathrm{51840}\mp\mathrm{23184}\sqrt{\mathrm{5}}\:} \\ $$$$=\frac{−\mathrm{10944}\sqrt{\mathrm{5}}}{\left(\mathrm{51840}\right)^{\mathrm{2}} −\left(\mathrm{23184}^{\mathrm{2}} .\mathrm{5}\right)} \\ $$$$=\frac{−\mathrm{10944}\sqrt{\mathrm{5}}}{−\mathrm{103680}} \\ $$$$=\frac{\mathrm{19}\sqrt{\mathrm{5}}\:}{\mathrm{180}} \\ $$

Commented by abdullah_ff last updated on 25/Nov/21

$$\mathrm{Rasheed}\:\mathrm{sir},\:\mathrm{you}\:\mathrm{are}\:{really}\:\mathrm{great}.. \\ $$

Answered by Rasheed.Sindhi last updated on 23/Nov/21

▶(x+(1/x))^2 =(2(√5) )^2 ⇒x^2 +(1/x^2 )=20−2 ⇒(x^2 +(1/x^2 ))^2 =(18)^2 ⇒x^4 +(1/x^4 )=322...(i) ⇒(x^4 +(1/x^4 ))^2 =(322)^2 ⇒x^8 +(1/x^8 )=322^2 −2...(ii) ▶(x+(1/x))^3 =(2(√5) )^3 ⇒x^3 +(1/x^3 )−3(x+(1/x))=8.5.(√5) ⇒x^3 +(1/x^3 )−3(2(√5) )=8.5.(√5) ⇒x^3 +(1/x^3 )=46(√5) ........(iii) (i)×(iii): (x^4 +(1/x^4 ))(x^3 +(1/x^3 ))=(322)(46(√5)) x^7 +(1/x^7 )+x+(1/x)=(322)(46(√5)) x^7 +(1/x^7 )+2(√5)=(322)(46(√5)) x^7 +(1/x^7 )=(322)(46(√5))−2(√5) ▶((x(x^6 − 1))/(x^8 − 1))=(x^7 /(x^8 −1))−(x/(x^8 −1)) =(1/((x^8 −1)/x^7 ))−(1/((x^8 −1)/x))=(1/(x−(1/x^7 )))−(1/(x^7 −(1/x))) =(((x^7 −(1/x))−(x−(1/x^7 )))/((x−(1/x^7 ))(x^7 −(1/x)))) =(((x^7 +(1/x^7 ))−(x+(1/x)))/((x^8 +(1/x^8 ))−2)) =((()−(x+(1/x)))/((x^8 +(1/x^8 ))−2))

$$\blacktriangleright\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\left(\mathrm{2}\sqrt{\mathrm{5}}\:\right)^{\mathrm{2}} \Rightarrow{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{20}−\mathrm{2} \\ $$$$\Rightarrow\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} =\left(\mathrm{18}\right)^{\mathrm{2}} \Rightarrow{x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }=\mathrm{322}…\left({i}\right) \\ $$$$ \\ $$$$ \\ $$$$\Rightarrow\left({x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)^{\mathrm{2}} =\left(\mathrm{322}\right)^{\mathrm{2}} \Rightarrow{x}^{\mathrm{8}} +\frac{\mathrm{1}}{{x}^{\mathrm{8}} }=\mathrm{322}^{\mathrm{2}} −\mathrm{2}…\left({ii}\right) \\ $$$$ \\ $$$$ \\ $$$$\blacktriangleright\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} =\left(\mathrm{2}\sqrt{\mathrm{5}}\:\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }−\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right)=\mathrm{8}.\mathrm{5}.\sqrt{\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }−\mathrm{3}\left(\mathrm{2}\sqrt{\mathrm{5}}\:\right)=\mathrm{8}.\mathrm{5}.\sqrt{\mathrm{5}} \\ $$$$\:\:\:\:\:\:\Rightarrow{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\mathrm{46}\sqrt{\mathrm{5}}\:……..\left({iii}\right) \\ $$$$ \\ $$$$ \\ $$$$\left({i}\right)×\left({iii}\right): \\ $$$$\left({x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)=\left(\mathrm{322}\right)\left(\mathrm{46}\sqrt{\mathrm{5}}\right) \\ $$$${x}^{\mathrm{7}} +\frac{\mathrm{1}}{{x}^{\mathrm{7}} }+{x}+\frac{\mathrm{1}}{{x}}=\left(\mathrm{322}\right)\left(\mathrm{46}\sqrt{\mathrm{5}}\right) \\ $$$${x}^{\mathrm{7}} +\frac{\mathrm{1}}{{x}^{\mathrm{7}} }+\mathrm{2}\sqrt{\mathrm{5}}=\left(\mathrm{322}\right)\left(\mathrm{46}\sqrt{\mathrm{5}}\right) \\ $$$${x}^{\mathrm{7}} +\frac{\mathrm{1}}{{x}^{\mathrm{7}} }=\left(\mathrm{322}\right)\left(\mathrm{46}\sqrt{\mathrm{5}}\right)−\mathrm{2}\sqrt{\mathrm{5}} \\ $$$$ \\ $$$$ \\ $$$$\blacktriangleright\frac{{x}\left({x}^{\mathrm{6}} \:−\:\mathrm{1}\right)}{{x}^{\mathrm{8}} \:−\:\mathrm{1}}=\frac{{x}^{\mathrm{7}} }{{x}^{\mathrm{8}} −\mathrm{1}}−\frac{{x}}{{x}^{\mathrm{8}} −\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\frac{{x}^{\mathrm{8}} −\mathrm{1}}{{x}^{\mathrm{7}} }}−\frac{\mathrm{1}}{\frac{{x}^{\mathrm{8}} −\mathrm{1}}{{x}}}=\frac{\mathrm{1}}{{x}−\frac{\mathrm{1}}{{x}^{\mathrm{7}} }}−\frac{\mathrm{1}}{{x}^{\mathrm{7}} −\frac{\mathrm{1}}{{x}}} \\ $$$$=\frac{\left({x}^{\mathrm{7}} −\frac{\mathrm{1}}{{x}}\right)−\left({x}−\frac{\mathrm{1}}{{x}^{\mathrm{7}} }\right)}{\left({x}−\frac{\mathrm{1}}{{x}^{\mathrm{7}} }\right)\left({x}^{\mathrm{7}} −\frac{\mathrm{1}}{{x}}\right)} \\ $$$$=\frac{\left({x}^{\mathrm{7}} +\frac{\mathrm{1}}{{x}^{\mathrm{7}} }\right)−\left({x}+\frac{\mathrm{1}}{{x}}\right)}{\left({x}^{\mathrm{8}} +\frac{\mathrm{1}}{{x}^{\mathrm{8}} }\right)−\mathrm{2}} \\ $$$$=\frac{\left(\right)−\left({x}+\frac{\mathrm{1}}{{x}}\right)}{\left({x}^{\mathrm{8}} +\frac{\mathrm{1}}{{x}^{\mathrm{8}} }\right)−\mathrm{2}} \\ $$

Commented by abdullah_ff last updated on 23/Nov/21

$${E}^{\mathcal{X}} \mathcal{C}\boldsymbol{\mathrm{e}}{lle}\boldsymbol{{nt}}\:\mathbb{S}\mathrm{i}\mathbb{R} \\ $$

Answered by som(math1967) last updated on 23/Nov/21

((x(x^6 −1))/(x^8 −1)) =(((x(x^6 −1))/x^4 )/((x^8 −1)/x^4 )) =((x^3 −(1/x^3 ))/(x^4 −(1/x^4 ))) =(((x−(1/x))(x^2 +1+(1/x^2 )))/((x−(1/x))(x+(1/x))(x^2 +(1/x^2 )))) [(x−(1/x))≠0] =(((x+(1/x))^2 −2.x.(1/x)+1)/(2(√5){(x+(1/x))^2 −2.x.(1/x)})) =((20−1)/(2(√5)(20−2))) =((19)/(36(√5)))=((19(√5))/(180))

$$\frac{{x}\left({x}^{\mathrm{6}} −\mathrm{1}\right)}{{x}^{\mathrm{8}} −\mathrm{1}} \\ $$$$=\frac{\frac{{x}\left({x}^{\mathrm{6}} −\mathrm{1}\right)}{{x}^{\mathrm{4}} }}{\frac{{x}^{\mathrm{8}} −\mathrm{1}}{{x}^{\mathrm{4}} }} \\ $$$$=\frac{{x}^{\mathrm{3}} −\frac{\mathrm{1}}{{x}^{\mathrm{3}} }}{{x}^{\mathrm{4}} −\frac{\mathrm{1}}{{x}^{\mathrm{4}} }} \\ $$$$=\frac{\cancel{\left({x}−\frac{\mathrm{1}}{{x}}\right)}\left({x}^{\mathrm{2}} +\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}{\cancel{\left({x}−\frac{\mathrm{1}}{{x}}\right)}\left({x}+\frac{\mathrm{1}}{{x}}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}\:\left[\left({x}−\frac{\mathrm{1}}{{x}}\right)\neq\mathrm{0}\right] \\ $$$$=\frac{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}.{x}.\frac{\mathrm{1}}{{x}}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}\left\{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}.{x}.\frac{\mathrm{1}}{{x}}\right\}} \\ $$$$=\frac{\mathrm{20}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}\left(\mathrm{20}−\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{19}}{\mathrm{36}\sqrt{\mathrm{5}}}=\frac{\mathrm{19}\sqrt{\mathrm{5}}}{\mathrm{180}} \\ $$

Commented by abdullah_ff last updated on 23/Nov/21

$$\mathrm{this}\:\mathrm{is}\:\mathrm{great}.\:\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}.. \\ $$

Commented by Rasheed.Sindhi last updated on 23/Nov/21

$$\mathcal{E}{fficient}! \\ $$

Answered by Tokugami last updated on 23/Nov/21

x+(1/x)=((x^2 +1)/x)=2(√5) (x+(1/x))^2 =(2(√5))^2 x^2 +2+(1/x^2 )=20 ((1+x^4 )/x^2 )=18 ((x(x^2 −1)(x^4 +x^2 +1))/((x^2 −1)(x^2 +1)(x^4 +1)))=((x/(x^2 +1)))(((x^4 +1+x^2 )/(x^4 +1)))=((x/(x^2 +1)))(1+(x^2 /(x^4 +1))) =((1/(2(√5))))(1+(1/(18))) =((19)/(36(√5)))

$${x}+\frac{\mathrm{1}}{{x}}=\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}}=\mathrm{2}\sqrt{\mathrm{5}} \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\left(\mathrm{2}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{20} \\ $$$$\frac{\mathrm{1}+{x}^{\mathrm{4}} }{{x}^{\mathrm{2}} }=\mathrm{18} \\ $$$$\frac{{x}\cancel{\left({x}^{\mathrm{2}} −\mathrm{1}\right)}\left({x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}\right)}{\cancel{\left({x}^{\mathrm{2}} −\mathrm{1}\right)}\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{4}} +\mathrm{1}\right)}=\left(\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\right)\left(\frac{{x}^{\mathrm{4}} +\mathrm{1}+{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} +\mathrm{1}}\right)=\left(\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\right)\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} +\mathrm{1}}\right) \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{18}}\right) \\ $$$$=\frac{\mathrm{19}}{\mathrm{36}\sqrt{\mathrm{5}}} \\ $$$$ \\ $$

Commented by abdullah_ff last updated on 23/Nov/21

$$\boldsymbol{{thanks}}\:\mathrm{for}\:\mathrm{your}\:\boldsymbol{{kind}}\:\mathrm{help}\:\boldsymbol{{sir}} \\ $$

Commented by Rasheed.Sindhi last updated on 23/Nov/21

$$\mathbb{E}\mathrm{fficient}! \\ $$

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