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](Q94638.png)
$$\alpha+\beta+\gamma=\mathrm{6}...........................\left({i}\right) \\ $$$$\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{3}} =\mathrm{87}......................\left({ii}\right) \\ $$$$\left(\alpha+\mathrm{1}\right)\left(\beta+\mathrm{1}\right)\left(\gamma+\mathrm{1}\right)=\mathrm{33}.........\left({iii}\right) \\ $$$$\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}+\frac{\mathrm{1}}{\gamma}=\frac{{m}}{{n}}\:\:\left[\:\:{gcd}\left({m},{n}\right)=\mathrm{1}\right] \\ $$$${m}+{n}=? \\ $$$$\vee\wedge\vee\wedge\vee\wedge\vee\wedge\vee\wedge\vee\wedge\vee\wedge\vee\wedge\vee\wedge\vee \\ $$$$\frac{{m}}{{n}}=\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}+\frac{\mathrm{1}}{\gamma}=\frac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}=? \\ $$$${So}\:{we}\:{need}\:{two}\:{quantities}:\: \\ $$$$\alpha\beta\gamma\:\&\:\alpha\beta+\beta\gamma+\gamma\alpha \\ $$$$\:^{\bullet} {Let}'{s}\:{use}\:\:{formula}\:{to}\:{determine} \\ $$$${a}\:{relation}\:{between}\:{these}\:{quantities}: \\ $$$$\left(\alpha+\beta+\gamma\right)\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} −\alpha\beta−\beta\gamma−\gamma\alpha\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{3}} −\mathrm{3}\alpha\beta\gamma \\ $$$$\left(\alpha+\beta+\gamma\right)\left\{\left(\alpha+\beta+\gamma\right)^{\mathrm{2}} −\mathrm{3}\alpha\beta−\mathrm{3}\beta\gamma−\mathrm{3}\gamma\alpha\right\} \\ $$$$\:\:\:\:\:\:\:\:\:=\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{3}} −\mathrm{3}\alpha\beta\gamma \\ $$$$\:\:\:\left(\mathrm{6}\right)\left(\mathrm{6}^{\mathrm{2}} −\mathrm{3}\left(\alpha\beta+\beta\gamma+\gamma\alpha\right)\right)=\mathrm{87}−\mathrm{3}\alpha\beta\gamma \\ $$$$\:\:\:\:\:\mathrm{72}−\mathrm{6}\left(\alpha\beta+\beta\gamma+\gamma\alpha\right)=\mathrm{29}−\alpha\beta\gamma \\ $$$$\alpha\beta\gamma−\mathrm{6}\left(\alpha\beta+\beta\gamma+\gamma\alpha\right)=−\mathrm{43} \\ $$$${let}\:\alpha\beta\gamma=\mathrm{p}\:\&\:\alpha\beta+\beta\gamma+\gamma\alpha=\mathrm{q} \\ $$$$\mathrm{p}−\mathrm{6q}=−\mathrm{43}.........................\mathrm{A} \\ $$$$\:^{\bullet} {Another}\:{relation}\:{can}\:{be}\:{derived} \\ $$$${from}\:\left({iii}\right) \\ $$$$\left({iii}\right)\Rightarrow\alpha\beta\gamma+\left(\alpha\beta+\beta\gamma+\gamma\alpha\right)+\left(\alpha+\beta+\gamma\right)+\mathrm{1}=\mathrm{33} \\ $$$$\:\alpha\beta\gamma+\left(\alpha\beta+\beta\gamma+\gamma\alpha\right)=\mathrm{26} \\ $$$$\:\:\:\mathrm{p}+\mathrm{q}=\mathrm{26}............................\mathrm{B} \\ $$$$\:^{\bullet\:} {From}\:\mathrm{A}\:\&\:\mathrm{B}: \\ $$$$−\mathrm{7q}=−\mathrm{69} \\ $$$$\:\:\:\:\:\:\:\mathrm{q}=\frac{\mathrm{69}}{\mathrm{7}} \\ $$$$\:\:\:\:\:\:\mathrm{p}=\mathrm{26}−\mathrm{q}=\mathrm{26}−\frac{\mathrm{69}}{\mathrm{7}}=\frac{\mathrm{182}−\mathrm{69}}{\mathrm{7}}=\frac{\mathrm{113}}{\mathrm{7}} \\ $$$$\:\:\:\frac{\mathrm{q}}{\mathrm{p}}=\frac{\mathrm{113}/\mathrm{7}}{\mathrm{69}/\mathrm{7}}=\frac{\mathrm{113}}{\mathrm{69}}=\frac{{m}}{{n}} \\ $$$$\:^{{Note}\:{that}\:\mathrm{113}\:\&\:\mathrm{69}\:{are}\:{coprime}.} \\ $$$${Hence} \\ $$$${m}+{n}=\mathrm{113}+\mathrm{69}=\mathrm{182} \\ $$
Commented by I want to learn more last updated on 20/May/20
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$