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Question-210961

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Question Number 210961 by RojaTaniya last updated on 24/Aug/24
Answered by A5T last updated on 24/Aug/24
((a−1+2)/(a−1))+((b−1+2)/(b−1))+((c−1+2)/(c−1))=10 ⇒(1/(a−1))+(1/(b−1))+(1/(c−1))=(7/2) a+b+c=0;ab+bc+ca=((m−1)/m);abc=((−m−1)/m) ⇒((ab+bc+ca−2(a+b+c)+3)/(abc−ab−bc−ac+a+b+c−1))=(7/2) ⇒(((m−1+3m)/m)/((−m−1+1−m−m)/m))=(7/2)⇒((4m−1)/(−3m))=(7/2) ⇒m=(2/(29))
$$\frac{{a}−\mathrm{1}+\mathrm{2}}{{a}−\mathrm{1}}+\frac{{b}−\mathrm{1}+\mathrm{2}}{{b}−\mathrm{1}}+\frac{{c}−\mathrm{1}+\mathrm{2}}{{c}−\mathrm{1}}=\mathrm{10} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}−\mathrm{1}}+\frac{\mathrm{1}}{{b}−\mathrm{1}}+\frac{\mathrm{1}}{{c}−\mathrm{1}}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$${a}+{b}+{c}=\mathrm{0};{ab}+{bc}+{ca}=\frac{{m}−\mathrm{1}}{{m}};{abc}=\frac{−{m}−\mathrm{1}}{{m}} \\ $$$$\Rightarrow\frac{{ab}+{bc}+{ca}−\mathrm{2}\left({a}+{b}+{c}\right)+\mathrm{3}}{{abc}−{ab}−{bc}−{ac}+{a}+{b}+{c}−\mathrm{1}}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\frac{{m}−\mathrm{1}+\mathrm{3}{m}}{{m}}}{\frac{−{m}−\mathrm{1}+\mathrm{1}−{m}−{m}}{{m}}}=\frac{\mathrm{7}}{\mathrm{2}}\Rightarrow\frac{\mathrm{4}{m}−\mathrm{1}}{−\mathrm{3}{m}}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\Rightarrow{m}=\frac{\mathrm{2}}{\mathrm{29}} \\ $$
Answered by Frix last updated on 24/Aug/24
Slightly weird method but I like it... x^3 +((m−1)/m)x+((m+1)/m)=0 (x−a)(x−b)(x−c)=x^3 +px+q ⇒ a=α∧b, c =−(α/2)±(√β) ⇒ p=−(((3α^2 )/4)+β)=((m−1)/m) q=−((α^3 /4)−αβ)=((m+1)/m) ⇒ β=((α^3 +3α^2 +8)/(4(α−1)))∧m=((α−1)/(α^3 +α+1)) (Σ((x+1)/(x−1)))−10=0 [inserting β] α^3 −((27)/2)α+((31)/2)=0 ⇒ α^3 =((27α−31)/2) m=((α−1)/(α^3 +α+1))=(2/(29))
$$\mathrm{Slightly}\:\mathrm{weird}\:\mathrm{method}\:\mathrm{but}\:\mathrm{I}\:\mathrm{like}\:\mathrm{it}… \\ $$$${x}^{\mathrm{3}} +\frac{{m}−\mathrm{1}}{{m}}{x}+\frac{{m}+\mathrm{1}}{{m}}=\mathrm{0} \\ $$$$\left({x}−{a}\right)\left({x}−{b}\right)\left({x}−{c}\right)={x}^{\mathrm{3}} +{px}+{q} \\ $$$$\Rightarrow\:{a}=\alpha\wedge{b},\:{c}\:=−\frac{\alpha}{\mathrm{2}}\pm\sqrt{\beta} \\ $$$$\Rightarrow \\ $$$${p}=−\left(\frac{\mathrm{3}\alpha^{\mathrm{2}} }{\mathrm{4}}+\beta\right)=\frac{{m}−\mathrm{1}}{{m}} \\ $$$${q}=−\left(\frac{\alpha^{\mathrm{3}} }{\mathrm{4}}−\alpha\beta\right)=\frac{{m}+\mathrm{1}}{{m}} \\ $$$$\Rightarrow\:\beta=\frac{\alpha^{\mathrm{3}} +\mathrm{3}\alpha^{\mathrm{2}} +\mathrm{8}}{\mathrm{4}\left(\alpha−\mathrm{1}\right)}\wedge{m}=\frac{\alpha−\mathrm{1}}{\alpha^{\mathrm{3}} +\alpha+\mathrm{1}} \\ $$$$\left(\Sigma\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)−\mathrm{10}=\mathrm{0}\:\:\:\:\:\left[\mathrm{inserting}\:\beta\right] \\ $$$$\alpha^{\mathrm{3}} −\frac{\mathrm{27}}{\mathrm{2}}\alpha+\frac{\mathrm{31}}{\mathrm{2}}=\mathrm{0}\:\Rightarrow\:\alpha^{\mathrm{3}} =\frac{\mathrm{27}\alpha−\mathrm{31}}{\mathrm{2}} \\ $$$${m}=\frac{\alpha−\mathrm{1}}{\alpha^{\mathrm{3}} +\alpha+\mathrm{1}}=\frac{\mathrm{2}}{\mathrm{29}} \\ $$
Answered by mr W last updated on 24/Aug/24
((a+1)/(a−1))+((b+1)/(b−1))+((c+1)/(c−1))=10 3+2((1/(a−1))+(1/(b−1))+(1/(c−1)))=10 ⇒(1/(a−1))+(1/(b−1))+(1/(c−1))=Σ(1/(x−1))=(7/2) mx^3 +(m−1)x+m+1=0 m(x−1+1)^3 +(m−1)(x−1+1)+m+1=0 with t=x−1 m(t+1)^3 +(m−1)(t+1)+m+1=0 mt^3 +3mt^2 +(4m−1)t+3m=0 m+((3m)/t)+((4m−1)/t^2 )+((3m)/t^3 )=0 ⇒Σ(1/t)=−((4m−1)/(3m))=Σ(1/(x−1))=(7/2) ⇒m=(2/(29)) ✓
$$\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}+\frac{{b}+\mathrm{1}}{{b}−\mathrm{1}}+\frac{{c}+\mathrm{1}}{{c}−\mathrm{1}}=\mathrm{10} \\ $$$$\mathrm{3}+\mathrm{2}\left(\frac{\mathrm{1}}{{a}−\mathrm{1}}+\frac{\mathrm{1}}{{b}−\mathrm{1}}+\frac{\mathrm{1}}{{c}−\mathrm{1}}\right)=\mathrm{10} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}−\mathrm{1}}+\frac{\mathrm{1}}{{b}−\mathrm{1}}+\frac{\mathrm{1}}{{c}−\mathrm{1}}=\Sigma\frac{\mathrm{1}}{{x}−\mathrm{1}}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$${mx}^{\mathrm{3}} +\left({m}−\mathrm{1}\right){x}+{m}+\mathrm{1}=\mathrm{0} \\ $$$${m}\left({x}−\mathrm{1}+\mathrm{1}\right)^{\mathrm{3}} +\left({m}−\mathrm{1}\right)\left({x}−\mathrm{1}+\mathrm{1}\right)+{m}+\mathrm{1}=\mathrm{0} \\ $$$${with}\:{t}={x}−\mathrm{1} \\ $$$${m}\left({t}+\mathrm{1}\right)^{\mathrm{3}} +\left({m}−\mathrm{1}\right)\left({t}+\mathrm{1}\right)+{m}+\mathrm{1}=\mathrm{0} \\ $$$${mt}^{\mathrm{3}} +\mathrm{3}{mt}^{\mathrm{2}} +\left(\mathrm{4}{m}−\mathrm{1}\right){t}+\mathrm{3}{m}=\mathrm{0} \\ $$$${m}+\frac{\mathrm{3}{m}}{{t}}+\frac{\mathrm{4}{m}−\mathrm{1}}{{t}^{\mathrm{2}} }+\frac{\mathrm{3}{m}}{{t}^{\mathrm{3}} }=\mathrm{0} \\ $$$$\Rightarrow\Sigma\frac{\mathrm{1}}{{t}}=−\frac{\mathrm{4}{m}−\mathrm{1}}{\mathrm{3}{m}}=\Sigma\frac{\mathrm{1}}{{x}−\mathrm{1}}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\Rightarrow{m}=\frac{\mathrm{2}}{\mathrm{29}}\:\checkmark \\ $$

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