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Question Number 147116 by Rasheed.Sindhi last updated on 18/Jul/21
a+b+c=1 a^2 +b^2 +c^2 =1 a^3 +b^3 +c^3 =1 a , b , c =?
$${a}+{b}+{c}=\mathrm{1} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{1} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{1} \\ $$$${a}\:,\:{b}\:,\:{c}\:=? \\ $$
Commented by mr W last updated on 18/Jul/21
(a,b,c)=(0,0,1) we can get ab+bc+ca=0, abc=0, therefore a,b,c are roots of x^3 −x^2 +0x+0=0 ⇒x^2 (x−1)=0 ⇒x∈(0,0,1)
$$\left({a},{b},{c}\right)=\left(\mathrm{0},\mathrm{0},\mathrm{1}\right) \\ $$$${we}\:{can}\:{get}\:{ab}+{bc}+{ca}=\mathrm{0},\:{abc}=\mathrm{0}, \\ $$$${therefore}\:{a},{b},{c}\:{are}\:{roots}\:{of} \\ $$$${x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{0}{x}+\mathrm{0}=\mathrm{0}\:\Rightarrow{x}^{\mathrm{2}} \left({x}−\mathrm{1}\right)=\mathrm{0}\: \\ $$$$\Rightarrow{x}\in\left(\mathrm{0},\mathrm{0},\mathrm{1}\right) \\ $$
Commented by Rasheed.Sindhi last updated on 18/Jul/21
Sir I ′ve tried to solve in a different way in which 2 also appears as a root. I know this can′t satisfy the system.But then I can′t deside whether it′s extraneous root because I haven′t squared any equation. Please see my solution and guide me as you always do.
$$\mathcal{S}{ir}\:{I}\:'{ve}\:{tried}\:{to}\:{solve}\:{in}\:{a}\:{different} \\ $$$${way}\:{in}\:{which}\:\mathrm{2}\:{also}\:{appears}\:{as}\:{a} \\ $$$${root}.\:{I}\:{know}\:{this}\:{can}'{t}\:{satisfy}\:{the} \\ $$$${system}.{But}\:{then}\:{I}\:{can}'{t}\:{deside}\: \\ $$$${whether}\:{it}'{s}\:{extraneous}\:{root}\:{because} \\ $$$${I}\:{haven}'{t}\:{squared}\:{any}\:{equation}. \\ $$$${Please}\:{see}\:{my}\:{solution}\:{and}\:{guide} \\ $$$${me}\:{as}\:\:{you}\:{always}\:{do}. \\ $$
Commented by Rasheed.Sindhi last updated on 18/Jul/21
Sorry sir I found my error when I solved it this time! Thanks Sir.
$${Sorry}\:{sir}\:{I}\:{found}\:{my}\:{error}\:{when} \\ $$$${I}\:{solved}\:{it}\:{this}\:{time}! \\ $$$$\mathcal{T}{hanks}\:\mathcal{S}{ir}. \\ $$
Commented by mr W last updated on 18/Jul/21
i think you also get the right answer in that way. you get c=1 then a+b=0 and ab=0, i.e. a=b=0. or c=0 then a+b=1 and ab=0, i.e. a=1, b=0 or a=0, b=1.
$${i}\:{think}\:{you}\:{also}\:{get}\:{the}\:{right}\:{answer} \\ $$$${in}\:{that}\:{way}.\:{you}\:{get}\:{c}=\mathrm{1}\:{then}\:{a}+{b}=\mathrm{0} \\ $$$${and}\:{ab}=\mathrm{0},\:{i}.{e}.\:{a}={b}=\mathrm{0}.\:{or}\:{c}=\mathrm{0}\:{then} \\ $$$${a}+{b}=\mathrm{1}\:{and}\:{ab}=\mathrm{0},\:{i}.{e}.\:{a}=\mathrm{1},\:{b}=\mathrm{0}\:{or} \\ $$$${a}=\mathrm{0},\:{b}=\mathrm{1}. \\ $$
Commented by Rasheed.Sindhi last updated on 18/Jul/21
ThanX for this nice trick Sir, I didn′t think of that.
$$\mathcal{T}{han}\mathcal{X}\:{for}\:{this}\:{nice}\:{trick}\:\mathcal{S}{ir}, \\ $$$${I}\:{didn}'{t}\:{think}\:{of}\:{that}. \\ $$
Answered by Olaf_Thorendsen last updated on 18/Jul/21
a+b+c = 1 a^2 +b^2 +c^2 = 1 a^3 +b^3 +c^3 = 1 a^2 +b^2 +c^2 = (a+b+c)^2 −2(ab+bc+ca) 1 = 1−2(ab+bc+ca) ab+bc+ca = 0 a^3 +b^3 +c^3 = (a+b+c)(a^2 +b^2 +c^2 ) −(ab^2 +ac^2 +a^2 b+bc^2 +a^2 c+b^2 c) 1 = 1×1−(ab^2 +ac^2 +a^2 b+bc^2 +a^2 c+b^2 c) 0 = a(ab+ca)+b(ab+bc)+c(bc+ca) 0 = a(−bc)+b(−ca)+c(−ab) abc = 0 we verify that a = 0 ⇒ b = 0 and c = 1 or b = 1 and c = 0 S = {(1,0,0) ; (0,1,0) ; (0,0,1)}
$${a}+{b}+{c}\:=\:\mathrm{1} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:=\:\mathrm{1} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \:=\:\mathrm{1} \\ $$$$ \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:=\:\left({a}+{b}+{c}\right)^{\mathrm{2}} −\mathrm{2}\left({ab}+{bc}+{ca}\right) \\ $$$$\mathrm{1}\:=\:\mathrm{1}−\mathrm{2}\left({ab}+{bc}+{ca}\right) \\ $$$${ab}+{bc}+{ca}\:=\:\mathrm{0} \\ $$$$ \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \:=\:\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right) \\ $$$$−\left({ab}^{\mathrm{2}} +{ac}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}+{bc}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}+{b}^{\mathrm{2}} {c}\right) \\ $$$$ \\ $$$$\mathrm{1}\:=\:\mathrm{1}×\mathrm{1}−\left({ab}^{\mathrm{2}} +{ac}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}+{bc}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}+{b}^{\mathrm{2}} {c}\right) \\ $$$$ \\ $$$$\mathrm{0}\:=\:{a}\left({ab}+{ca}\right)+{b}\left({ab}+{bc}\right)+{c}\left({bc}+{ca}\right) \\ $$$$\mathrm{0}\:=\:{a}\left(−{bc}\right)+{b}\left(−{ca}\right)+{c}\left(−{ab}\right) \\ $$$${abc}\:=\:\mathrm{0} \\ $$$${we}\:{verify}\:{that}\:{a}\:=\:\mathrm{0}\:\Rightarrow\:{b}\:=\:\mathrm{0}\:\mathrm{and}\:{c}\:=\:\mathrm{1} \\ $$$$\mathrm{or}\:{b}\:=\:\mathrm{1}\:\mathrm{and}\:{c}\:=\:\mathrm{0} \\ $$$$ \\ $$$$\mathcal{S}\:=\:\left\{\left(\mathrm{1},\mathrm{0},\mathrm{0}\right)\:;\:\left(\mathrm{0},\mathrm{1},\mathrm{0}\right)\:;\:\left(\mathrm{0},\mathrm{0},\mathrm{1}\right)\right\} \\ $$
Commented by Rasheed.Sindhi last updated on 18/Jul/21
Thanks Sir Olaf! I always appreciate your precious contribution towards this forum!
$$\mathcal{T}{hanks}\:\mathcal{S}{ir}\:\mathrm{Olaf}!\:{I}\:{always}\:{appreciate} \\ $$$${your}\:{precious}\:{contribution}\:{towards}\:{this} \\ $$$${forum}! \\ $$
Answered by Rasheed.Sindhi last updated on 18/Jul/21
a+b+c=a^2 +b^2 +c^2 =a^3 +b^3 +c^3 =1_(−) ;a,b,c=? (i)⇒a+b=1−c (ii)⇒(a+b)^2 −2ab+c^2 −1=0 (1−c)^2 −2ab+c^2 −1=0 ab=((1−2c+c^2 +c^2 −1)/2)=c^2 −c...A (iii)⇒(a+b)^3 −3ab(a+b)+c^3 −1=0 (1−c)^3 −3ab(1−c)−(1−c^3 )=0 (1−c){(1−c)^2 −3ab−(1+c+c^2 )}=0 c=1 ∨ 1−2c+c^2 −3ab−1−c−c^2 =0 ab=−c...........................B A & B : c^2 −c=−c c^2 =0⇒c=0 Observing symmetric nature of the system we can fix same values for a & b. • •^(•) (a,b,c)=(1,0,0) , (0,1,0) , (0,0,1)
$$\underset{−} {{a}+{b}+{c}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} ={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{1}};{a},{b},{c}=? \\ $$$$\left({i}\right)\Rightarrow{a}+{b}=\mathrm{1}−{c} \\ $$$$\left({ii}\right)\Rightarrow\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}{ab}+{c}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\left(\mathrm{1}−{c}\right)^{\mathrm{2}} −\mathrm{2}{ab}+{c}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:{ab}=\frac{\mathrm{1}−\mathrm{2}{c}+{c}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}}={c}^{\mathrm{2}} −{c}…{A} \\ $$$$\left({iii}\right)\Rightarrow\left({a}+{b}\right)^{\mathrm{3}} −\mathrm{3}{ab}\left({a}+{b}\right)+{c}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\left(\mathrm{1}−{c}\right)^{\mathrm{3}} −\mathrm{3}{ab}\left(\mathrm{1}−{c}\right)−\left(\mathrm{1}−{c}^{\mathrm{3}} \right)=\mathrm{0} \\ $$$$\:\left(\mathrm{1}−{c}\right)\left\{\left(\mathrm{1}−{c}\right)^{\mathrm{2}} −\mathrm{3}{ab}−\left(\mathrm{1}+{c}+{c}^{\mathrm{2}} \right)\right\}=\mathrm{0} \\ $$$${c}=\mathrm{1}\:\vee\:\cancel{\mathrm{1}}−\mathrm{2}{c}+\cancel{{c}^{\mathrm{2}} }−\mathrm{3}{ab}−\cancel{\mathrm{1}}−{c}−\cancel{{c}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ab}=−{c}………………………{B} \\ $$$${A}\:\&\:{B}\::\:\:\:{c}^{\mathrm{2}} −{c}=−{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{c}^{\mathrm{2}} =\mathrm{0}\Rightarrow{c}=\mathrm{0} \\ $$$${Observing}\:{symmetric}\:{nature}\:{of} \\ $$$${the}\:{system}\:{we}\:{can}\:{fix}\:{same}\:{values} \\ $$$${for}\:{a}\:\&\:{b}. \\ $$$$\overset{\bullet} {\bullet\:\:\:\bullet}\:\left({a},{b},{c}\right)=\left(\mathrm{1},\mathrm{0},\mathrm{0}\right)\:,\:\left(\mathrm{0},\mathrm{1},\mathrm{0}\right)\:,\:\left(\mathrm{0},\mathrm{0},\mathrm{1}\right) \\ $$

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