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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)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+γα∙Let′suseformulatodeterminearelationbetweenthesequantities:(α+β+γ)(α2+β2+γ2−αβ−βγ−γα)=α3+β3+γ3−3αβγ(α+β+γ){(α+β+γ)2−3αβ−3βγ−3γα}=α3+β3+γ3−3αβγ(6)(62−3(αβ+βγ+γα))=87−3αβγ72−6(αβ+βγ+γα)=29−αβγαβγ−6(αβ+βγ+γα)=−43letαβγ=p&αβ+βγ+γα=qp−6q=−43.........................A∙Anotherrelationcanbederivedfrom(iii)(iii)⇒αβγ+(αβ+βγ+γα)+(α+β+γ)+1=33αβγ+(αβ+βγ+γα)=26p+q=26............................B∙FromA&B:−7q=−69q=697p=26−q=26−697=182−697=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
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