Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com