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Question Number 94601 by I want to learn more last updated on 19/May/20

Answered by Rasheed.Sindhi last updated on 20/May/20

α+β+γ=6...........................(i)  α^3 +β^3 +γ^3 =87......................(ii)  (α+1)(β+1)(γ+1)=33.........(iii)  (1/α)+(1/β)+(1/γ)=(m/n)  [  gcd(m,n)=1]  m+n=?  ∨∧∨∧∨∧∨∧∨∧∨∧∨∧∨∧∨∧∨  (m/n)=(1/α)+(1/β)+(1/γ)=((αβ+βγ+γα)/(αβγ))=?  So we need two quantities:   αβγ & αβ+βγ+γα  ^• Let′s use  formula to determine  a relation between these quantities:  (α+β+γ)(α^2 +β^2 +γ^2 −αβ−βγ−γα)           =α^3 +β^3 +γ^3 −3αβγ  (α+β+γ){(α+β+γ)^2 −3αβ−3βγ−3γα}           =α^3 +β^3 +γ^3 −3αβγ     (6)(6^2 −3(αβ+βγ+γα))=87−3αβγ       72−6(αβ+βγ+γα)=29−αβγ  αβγ−6(αβ+βγ+γα)=−43  let αβγ=p & αβ+βγ+γα=q  p−6q=−43.........................A  ^• Another relation can be derived  from (iii)  (iii)⇒αβγ+(αβ+βγ+γα)+(α+β+γ)+1=33   αβγ+(αβ+βγ+γα)=26     p+q=26............................B  ^(• ) From A & B:  −7q=−69         q=((69)/7)        p=26−q=26−((69)/7)=((182−69)/7)=((113)/7)     (q/p)=((113/7)/(69/7))=((113)/(69))=(m/n)  ^(Note that 113 & 69 are coprime.)   Hence  m+n=113+69=182

α+β+γ=6...........................(i)α3+β3+γ3=87......................(ii)(α+1)(β+1)(γ+1)=33.........(iii)1α+1β+1γ=mn[gcd(m,n)=1]m+n=?mn=1α+1β+1γ=αβ+βγ+γααβγ=?Soweneedtwoquantities:αβγ&αβ+βγ+γαLetsuseformulatodeterminearelationbetweenthesequantities:(α+β+γ)(α2+β2+γ2αββγγα)=α3+β3+γ33αβγ(α+β+γ){(α+β+γ)23αβ3βγ3γα}=α3+β3+γ33αβγ(6)(623(αβ+βγ+γα))=873αβγ726(αβ+βγ+γα)=29αβγαβγ6(αβ+βγ+γα)=43letαβγ=p&αβ+βγ+γα=qp6q=43.........................AAnotherrelationcanbederivedfrom(iii)(iii)αβγ+(αβ+βγ+γα)+(α+β+γ)+1=33αβγ+(αβ+βγ+γα)=26p+q=26............................BFromA&B:7q=69q=697p=26q=26697=182697=1137qp=113/769/7=11369=mnNotethat113&69arecoprime.Hencem+n=113+69=182

Commented by I want to learn more last updated on 20/May/20

I really appreciate sir.

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