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x-3-1-x-3-1-then-prove-that-x-5-1-x-5-3-1-x-5-1-x-5-3-

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Question Number 174177 by mathlove last updated on 26/Jul/22
x^3 +(1/x^3 )=1 then prove that (((x^5 +(1/x^5 ))^3 −1)/(x^5 +(1/x^5 )))=3
x3+1x3=1thenprovethat(x5+1x5)3−1x5+1x5=3
Answered by mr W last updated on 26/Jul/22
(x+(1/x))^3 =x^3 +(1/x^3 )+3(x+(1/x)) P=x+(1/x) P^3 =3P+1 (x+(1/x))^2 =x^2 +(1/x^2 )+2 x^2 +(1/x^2 )=P^2 −2 (x^3 +(1/x^3 ))(x^2 +(1/x^2 ))=x^5 +(1/x^5 )+x+(1/x) x^5 +(1/x^5 )=P^2 −P−2 (x^5 +(1/x^5 ))^3 =P^6 −3P^5 −3P^4 +11P^3 +6P^2 −12P−8 (x^5 +(1/x^5 ))^3 =(3P+1)^2 −3P^2 (3P+1)−3P(3P+1)+11(3P+1)+6P^2 −12P−8 (x^5 +(1/x^5 ))^3 =9P^2 +6P+1−9P^3 −3P^2 −9P^2 −3P+33P+11+6P^2 −12P−8 (x^5 +(1/x^5 ))^3 =−9P^3 +3P^2 +24P+4 (x^5 +(1/x^5 ))^3 =−9(3P+1)+3P^2 +24P+4 (x^5 +(1/x^5 ))^3 =3P^2 −3P−5 (x^5 +(1/x^5 ))^3 −1=3P^2 −3P−6 (x^5 +(1/x^5 ))^3 −1=3(P^2 −P−2)=3(x^5 +(1/x^5 )) ⇒(((x^5 +(1/x^5 ))^3 −1)/((x^5 +(1/x^5 ))))=3 ✓
(x+1x)3=x3+1x3+3(x+1x)P=x+1xP3=3P+1(x+1x)2=x2+1x2+2x2+1x2=P2−2(x3+1x3)(x2+1x2)=x5+1x5+x+1xx5+1x5=P2−P−2(x5+1x5)3=P6−3P5−3P4+11P3+6P2−12P−8(x5+1x5)3=(3P+1)2−3P2(3P+1)−3P(3P+1)+11(3P+1)+6P2−12P−8(x5+1x5)3=9P2+6P+1−9P3−3P2−9P2−3P+33P+11+6P2−12P−8(x5+1x5)3=−9P3+3P2+24P+4(x5+1x5)3=−9(3P+1)+3P2+24P+4(x5+1x5)3=3P2−3P−5(x5+1x5)3−1=3P2−3P−6(x5+1x5)3−1=3(P2−P−2)=3(x5+1x5)⇒(x5+1x5)3−1(x5+1x5)=3✓
Commented by mathlove last updated on 26/Jul/22
thanks mr W
thanksmrW
Answered by chengulapetrom last updated on 26/Jul/22
(x+(1/x))^5 =x^5 +5x^4 ((1/x))+10x^3 ((1/x))^2 +10x^2 ((1/x))^3 +5x((1/x))^4 +((1/x))^5 =x^5 +(1/x^5 )+5(x^3 +(1/x^3 ))+10(x+(1/x)) =x^5 +(1/x^5 )+10(x+(1/x))+5 (x+(1/x))^5 =x^5 +(1/x^5 )+10(x+(1/x))+5 (x+(1/x))^3 =x^3 +3(x+(1/x))+(1/x^3 ) (x+(1/x))^3 =3(x+(1/x))+1 (x+(1/x))^3 −3(x+(1/x))−1=0 (x+(1/x))^3 −1=3(x+(1/x)) (((x+(1/x))^3 −1)/((x+(1/x))))=3 similarly (((x^5 +(1/x^5 ))^3 −1)/(x^5 +(1/x^5 )))=3
(x+1x)5=x5+5x4(1x)+10x3(1x)2+10x2(1x)3+5x(1x)4+(1x)5=x5+1x5+5(x3+1x3)+10(x+1x)=x5+1x5+10(x+1x)+5(x+1x)5=x5+1x5+10(x+1x)+5(x+1x)3=x3+3(x+1x)+1x3(x+1x)3=3(x+1x)+1(x+1x)3−3(x+1x)−1=0(x+1x)3−1=3(x+1x)(x+1x)3−1(x+1x)=3similarly(x5+1x5)3−1x5+1x5=3
Commented by chengulapetrom last updated on 26/Jul/22
Mr W am I correct
MrWamIcorrect
Commented by mr W last updated on 27/Jul/22
i don′t think it′s “similary” that (((x^5 +(1/x^5 ))^3 −1)/(x^5 +(1/x^5 )))=3 and (((x+(1/x))^3 −1)/((x+(1/x))))=3. you must show it. or maybe i didn′t understand you correctly. in fact (((x^3 +(1/x^3 ))^3 −1)/(x^3 +(1/x^3 )))≠3 and (((x^4 +(1/x^4 ))^3 −1)/(x^4 +(1/x^4 )))≠3 etc.
idon′tthinkit′s“similary″that(x5+1x5)3−1x5+1x5=3and(x+1x)3−1(x+1x)=3.youmustshowit.ormaybeididn′tunderstandyoucorrectly.infact(x3+1x3)3−1x3+1x3≠3and(x4+1x4)3−1x4+1x4≠3etc.
Answered by mr W last updated on 26/Jul/22
let P_n =x^n +(1/x^n ) P_1 =e_1 =p P_2 =e_1 P_1 −2e_2 =p^2 −2 P_n =e_1 P_(n−1) −e_2 P_(n−2) P_3 =e_1 P_2 −P_1 1=p(p^2 −2)−p p^3 −3p−1=0 P_4 =pP_3 −P_2 =p−p^2 +2=−p^2 +p+2 P_5 =pP_4 −P_3 =(−p^2 +p+2)p−1=−p^3 +p^2 +2p−1 P_5 =−3p−1+p^2 +2p−1=p^2 −p−2 ......
letPn=xn+1xnP1=e1=pP2=e1P1−2e2=p2−2Pn=e1Pn−1−e2Pn−2P3=e1P2−P11=p(p2−2)−pp3−3p−1=0P4=pP3−P2=p−p2+2=−p2+p+2P5=pP4−P3=(−p2+p+2)p−1=−p3+p2+2p−1P5=−3p−1+p2+2p−1=p2−p−2……
Answered by Rasheed.Sindhi last updated on 27/Jul/22
x^6 =x^3 −1 x^5 =((x^3 −1)/x) ▶x^5 +(1/x^5 )=((x^3 −1)/x)+(x/(x^3 −1)) =(((x^3 −1)^2 +x^2 )/(x(x^3 −1)))=((x^6 −2x^3 +1+x^2 )/(x(x^3 −1))) =((x^3 −1−2x^3 +1+x^2 )/(x(x^3 −1))) =((−x^3 +x^2 )/(x(x^3 −1)))=((−x(x−1))/((x−1)(x^2 +x+1))) =((−1)/((x^2 +x+1)/x))=((−1)/(x+(1/x)_() +1))=−(1/(p+1)) ▶(((x^5 +(1/x^5 ))^3 −1)/(x^5 +(1/x^5 )))=(((−(1/(p+1)))^3 −1)/(−(1/(p+1)))) =((((1/(p+1)))^3 +1)/(1/(p+1))) =((((1/(p+1))+1)( ((1/(p+1)))^2 −((1/(p+1)))+1))/(1/(p+1))) =(((p+2)/(p+1)))(((1−(p+1)+(p+1)^2 )/((p+1)^2 )))(p+1) =(((p+2)(1−p−1+p^2 +2p+1))/((p+1)^2 )) =(((p+2)(p^2 +p+1))/((p+1)^2 )) =(((p−1+3)(p^2 +p+1))/((p+1)^2 )) =(((p−1)(p^2 +p+1)+3(p^2 +p+1))/((p+1)^2 )) =((p^3 −1+3p^2 +3p+3)/((p+1)^2 )) =((p^3 +3p^2 +3p+2)/((p+1)^2 ))−3+3 =((p^3 +3p^2 +3p+2−3p^2 −6p−3)/(p^2 +2p+1))+3 =((p^3 −3p−1)/(p^2 +2p+1))+3 =(((x+(1/x))^3 −3(x+(1/x))−1)/((x+(1/x))^2 +2(x+(1/x))+1))+3 =((x^3 +(1/x^3 )+3(x+(1/x))−3(x+(1/x))−1)/((x+(1/x))^2 +2(x+(1/x))+1))+3 =((1−1)/((x+(1/x))^2 +2(x+(1/x))+1))+3 =3
x6=x3−1x5=x3−1x▸x5+1x5=x3−1x+xx3−1=(x3−1)2+x2x(x3−1)=x6−2x3+1+x2x(x3−1)=x3−1−2x3+1+x2x(x3−1)=−x3+x2x(x3−1)=−x(x−1)(x−1)(x2+x+1)=−1x2+x+1x=−1x+1x⏟+1=−1p+1▸(x5+1x5)3−1x5+1x5=(−1p+1)3−1−1p+1=(1p+1)3+11p+1=(1p+1+1)((1p+1)2−(1p+1)+1)1p+1=(p+2p+1)(1−(p+1)+(p+1)2(p+1)2)(p+1)=(p+2)(1−p−1+p2+2p+1)(p+1)2=(p+2)(p2+p+1)(p+1)2=(p−1+3)(p2+p+1)(p+1)2=(p−1)(p2+p+1)+3(p2+p+1)(p+1)2=p3−1+3p2+3p+3(p+1)2=p3+3p2+3p+2(p+1)2−3+3=p3+3p2+3p+2−3p2−6p−3p2+2p+1+3=p3−3p−1p2+2p+1+3=(x+1x)3−3(x+1x)−1(x+1x)2+2(x+1x)+1+3=x3+1x3+3(x+1x)−3(x+1x)−1(x+1x)2+2(x+1x)+1+3=1−1(x+1x)2+2(x+1x)+1+3=3

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