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Question Number 83721 by mr W last updated on 05/Mar/20

Commented by MJS last updated on 05/Mar/20

−3

$$−\mathrm{3} \\ $$

Commented by john santu last updated on 05/Mar/20

(1)+(2)+(3) a+b+c−((1/a)+(1/b)+(1/c))= a+b+c (1/a)+(1/b)+(1/c) = 0 ((a+b)/(ab))+(1/c) = 0 ((ac+bc+ab)/(abc)) = 0 ⇒ ((c^2 −1+b^2 −1+a^2 −1)/(abc))=0 a^2 +b^2 +c^2 =3 (a+b+c)^2 −2(ab+ac+bc) = 3 a+b+c = (√3)

$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)+\left(\mathrm{3}\right) \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{c}−\left(\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}+\frac{\mathrm{1}}{\mathrm{c}}\right)=\:\mathrm{a}+\mathrm{b}+\mathrm{c} \\ $$$$\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}+\frac{\mathrm{1}}{\mathrm{c}}\:=\:\mathrm{0} \\ $$$$\frac{\mathrm{a}+\mathrm{b}}{\mathrm{ab}}+\frac{\mathrm{1}}{\mathrm{c}}\:=\:\mathrm{0} \\ $$$$\frac{\mathrm{ac}+\mathrm{bc}+\mathrm{ab}}{\mathrm{abc}}\:=\:\mathrm{0}\:\Rightarrow\:\frac{\mathrm{c}^{\mathrm{2}} −\mathrm{1}+\mathrm{b}^{\mathrm{2}} −\mathrm{1}+\mathrm{a}^{\mathrm{2}} −\mathrm{1}}{\mathrm{abc}}=\mathrm{0} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} =\mathrm{3} \\ $$$$\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{ab}+\mathrm{ac}+\mathrm{bc}\right)\:=\:\mathrm{3} \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{c}\:=\:\sqrt{\mathrm{3}} \\ $$$$ \\ $$

Commented by mr W last updated on 05/Mar/20

SORRY! −3 is the correct answer!

$$\boldsymbol{{SORRY}}! \\ $$$$−\mathrm{3}\:{is}\:{the}\:{correct}\:{answer}! \\ $$

Commented by john santu last updated on 05/Mar/20

haha..., i′m focus on finding the abc −value

$$\mathrm{haha}...,\:\mathrm{i}'\mathrm{m}\:\mathrm{focus}\:\mathrm{on}\:\mathrm{finding}\:\mathrm{the} \\ $$$$\mathrm{abc}\:−\mathrm{value} \\ $$

Commented by MJS last updated on 05/Mar/20

(1/a)+(1/b)+(1/c)=0 ⇒ c=−((ab)/(a+b)) (1/(ab))+(1/(bc))+(1/(ca))=0 ⇒ c=−(a+b) −((ab)/(a+b))=−(a+b) ⇒ b=a(−(1/2)±((√3)/2)i) ⇒ c=a(−(1/2)∓((√3)/2)i) b−(1/b)=c ⇒ a^2 =−(1/2)±((√3)/6)i but then a−(1/a)≠b ∧ c−(1/c)≠a

$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\mathrm{0}\:\Rightarrow\:{c}=−\frac{{ab}}{{a}+{b}} \\ $$$$\frac{\mathrm{1}}{{ab}}+\frac{\mathrm{1}}{{bc}}+\frac{\mathrm{1}}{{ca}}=\mathrm{0}\:\Rightarrow\:{c}=−\left({a}+{b}\right) \\ $$$$−\frac{{ab}}{{a}+{b}}=−\left({a}+{b}\right)\:\Rightarrow\:{b}={a}\left(−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right) \\ $$$$\Rightarrow\:{c}={a}\left(−\frac{\mathrm{1}}{\mathrm{2}}\mp\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right) \\ $$$${b}−\frac{\mathrm{1}}{{b}}={c} \\ $$$$\Rightarrow\:{a}^{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{6}}\mathrm{i} \\ $$$$\mathrm{but}\:\mathrm{then}\:{a}−\frac{\mathrm{1}}{{a}}\neq{b}\:\wedge\:{c}−\frac{\mathrm{1}}{{c}}\neq{a} \\ $$

Commented by Prithwish Sen 1 last updated on 05/Mar/20

But Sir, abc((1/a) + (1/b) + (1/c)) = ab + bc + ac

$$\mathrm{But}\:\mathrm{Sir}, \\ $$$$\mathrm{abc}\left(\frac{\mathrm{1}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}\:+\:\frac{\mathrm{1}}{\mathrm{c}}\right)\:=\:\mathrm{ab}\:+\:\mathrm{bc}\:+\:\mathrm{ac} \\ $$

Commented by MJS last updated on 05/Mar/20

see my solution below, it′s wrong?

$$\mathrm{see}\:\mathrm{my}\:\mathrm{solution}\:\mathrm{below},\:\mathrm{it}'\mathrm{s}\:\mathrm{wrong}? \\ $$

Commented by Prithwish Sen 1 last updated on 05/Mar/20

a−b = (1/a) ......(i), b−c = (1/b)....(ii), c−a=(1/c)...(iii) (i)×c + (ii)×a + (iii)×b we get (c/a) + (b/c) +(a/b) = 0 ....(iv) again , (i)×(1/b) + (ii)×(1/c) + (iii)×(1/a) we get (1/(ab))+(1/(bc))+(1/(ac)) = −3+((a/b)+(b/c)+(c/a)) = −3 (from iv)

$${a}−{b}\:=\:\frac{\mathrm{1}}{{a}}\:......\left({i}\right),\:{b}−{c}\:=\:\frac{\mathrm{1}}{{b}}....\left({ii}\right),\:{c}−{a}=\frac{\mathrm{1}}{{c}}...\left({iii}\right) \\ $$$$\left(\boldsymbol{\mathrm{i}}\right)×\boldsymbol{\mathrm{c}}\:+\:\left(\boldsymbol{\mathrm{ii}}\right)×\boldsymbol{\mathrm{a}}\:+\:\left(\boldsymbol{\mathrm{iii}}\right)×\boldsymbol{\mathrm{b}}\:\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{get}} \\ $$$$\frac{\boldsymbol{\mathrm{c}}}{\boldsymbol{\mathrm{a}}}\:+\:\frac{\boldsymbol{\mathrm{b}}}{\boldsymbol{\mathrm{c}}}\:+\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{b}}}\:=\:\mathrm{0}\:....\left(\boldsymbol{\mathrm{iv}}\right) \\ $$$$\boldsymbol{\mathrm{again}}\:,\:\left(\boldsymbol{\mathrm{i}}\right)×\frac{\mathrm{1}}{\boldsymbol{\mathrm{b}}}\:+\:\left(\boldsymbol{\mathrm{ii}}\right)×\frac{\mathrm{1}}{\boldsymbol{\mathrm{c}}}\:+\:\left(\boldsymbol{\mathrm{iii}}\right)×\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}}\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{get}} \\ $$$$\frac{\mathrm{1}}{\boldsymbol{\mathrm{ab}}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{bc}}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{ac}}}\:=\:−\mathrm{3}+\left(\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{b}}}+\frac{\boldsymbol{\mathrm{b}}}{\boldsymbol{\mathrm{c}}}+\frac{\boldsymbol{\mathrm{c}}}{\boldsymbol{\mathrm{a}}}\right)\:=\:−\mathrm{3}\:\:\left(\boldsymbol{\mathrm{from}}\:\boldsymbol{\mathrm{iv}}\right) \\ $$

Answered by MJS last updated on 05/Mar/20

b=((a^2 −1)/a) ⇒ c=((a^4 −3a^2 +1)/(a^3 −a)) ⇒ a=((a^8 −7a^6 +13a^4 −7a^2 +1)/(a^7 +4a^5 +4a^3 −a)) ((3a^6 −9a^4 +6a^2 −1)/(a^7 −4a^5 +4a^3 −a))=0 a^6 −3a^4 +2a^2 −(1/3)=0 y=a^2 y^3 −3y^2 +2y−(1/3)=0 z=y−1 z^3 −z−(1/3)=0 z_1 =((2(√3))/3)cos (π/(18)) z_2 =−((2(√3))/3)sin ((2π)/9) z_3 =−((2(√3))/3)sin (π/9) y_1 =1+((2(√3))/3)cos (π/(18)) y_2 =1−((2(√3))/3)sin ((2π)/9) y_3 =1−((2(√3))/3)sin (π/9) a_(1, 4) =±(√(1+((2(√3))/3)cos (π/(18)))) a_(2, 5) =±(√(1−((2(√3))/3)sin ((2π)/9))) a_(3, 6) =±(√(1−((2(√3))/3)sin (π/9))) with these values (1/(ab))+(1/(bc))+(1/(ca))=−3

$${b}=\frac{{a}^{\mathrm{2}} −\mathrm{1}}{{a}} \\ $$$$\Rightarrow\:{c}=\frac{{a}^{\mathrm{4}} −\mathrm{3}{a}^{\mathrm{2}} +\mathrm{1}}{{a}^{\mathrm{3}} −{a}} \\ $$$$\Rightarrow\:{a}=\frac{{a}^{\mathrm{8}} −\mathrm{7}{a}^{\mathrm{6}} +\mathrm{13}{a}^{\mathrm{4}} −\mathrm{7}{a}^{\mathrm{2}} +\mathrm{1}}{{a}^{\mathrm{7}} +\mathrm{4}{a}^{\mathrm{5}} +\mathrm{4}{a}^{\mathrm{3}} −{a}} \\ $$$$\frac{\mathrm{3}{a}^{\mathrm{6}} −\mathrm{9}{a}^{\mathrm{4}} +\mathrm{6}{a}^{\mathrm{2}} −\mathrm{1}}{{a}^{\mathrm{7}} −\mathrm{4}{a}^{\mathrm{5}} +\mathrm{4}{a}^{\mathrm{3}} −{a}}=\mathrm{0} \\ $$$${a}^{\mathrm{6}} −\mathrm{3}{a}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{0} \\ $$$${y}={a}^{\mathrm{2}} \\ $$$${y}^{\mathrm{3}} −\mathrm{3}{y}^{\mathrm{2}} +\mathrm{2}{y}−\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{0} \\ $$$${z}={y}−\mathrm{1} \\ $$$${z}^{\mathrm{3}} −{z}−\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{0} \\ $$$${z}_{\mathrm{1}} =\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{cos}\:\frac{\pi}{\mathrm{18}} \\ $$$${z}_{\mathrm{2}} =−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{9}} \\ $$$${z}_{\mathrm{3}} =−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{sin}\:\frac{\pi}{\mathrm{9}} \\ $$$${y}_{\mathrm{1}} =\mathrm{1}+\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{cos}\:\frac{\pi}{\mathrm{18}} \\ $$$${y}_{\mathrm{2}} =\mathrm{1}−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{9}} \\ $$$${y}_{\mathrm{3}} =\mathrm{1}−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{sin}\:\frac{\pi}{\mathrm{9}} \\ $$$${a}_{\mathrm{1},\:\mathrm{4}} =\pm\sqrt{\mathrm{1}+\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{cos}\:\frac{\pi}{\mathrm{18}}} \\ $$$${a}_{\mathrm{2},\:\mathrm{5}} =\pm\sqrt{\mathrm{1}−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{9}}} \\ $$$${a}_{\mathrm{3},\:\mathrm{6}} =\pm\sqrt{\mathrm{1}−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{sin}\:\frac{\pi}{\mathrm{9}}} \\ $$$$\mathrm{with}\:\mathrm{these}\:\mathrm{values} \\ $$$$\frac{\mathrm{1}}{{ab}}+\frac{\mathrm{1}}{{bc}}+\frac{\mathrm{1}}{{ca}}=−\mathrm{3} \\ $$

Commented by Prithwish Sen 1 last updated on 05/Mar/20

perfect sir . How r u sir .

$$\mathrm{perfect}\:\mathrm{sir}\:.\:\mathrm{How}\:\mathrm{r}\:\mathrm{u}\:\mathrm{sir}\:. \\ $$

Answered by mr W last updated on 05/Mar/20

Answered by behi83417@gmail.com last updated on 05/Mar/20

(a/b)−(1/(ab))=1,(b/c)−(1/(bc))=1,(c/a)−(1/(ac))=1 ⇒(1/(ab))+(1/(bc))+(1/(ca))=((a/b)+(b/c)+(c/a))−3=−3 [(a/b)+(b/c)+(c/a)=c(a−b)+a(b−c)+b(c−a)=0]

$$\frac{\mathrm{a}}{\mathrm{b}}−\frac{\mathrm{1}}{\mathrm{ab}}=\mathrm{1},\frac{\mathrm{b}}{\mathrm{c}}−\frac{\mathrm{1}}{\mathrm{bc}}=\mathrm{1},\frac{\mathrm{c}}{\mathrm{a}}−\frac{\mathrm{1}}{\mathrm{ac}}=\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{ab}}+\frac{\mathrm{1}}{\mathrm{bc}}+\frac{\mathrm{1}}{\mathrm{ca}}=\left(\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{a}}\right)−\mathrm{3}=−\mathrm{3} \\ $$$$\left[\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{a}}=\mathrm{c}\left(\mathrm{a}−\mathrm{b}\right)+\mathrm{a}\left(\mathrm{b}−\mathrm{c}\right)+\mathrm{b}\left(\mathrm{c}−\mathrm{a}\right)=\mathrm{0}\right] \\ $$

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