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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=1a2+b2+c2=1a3+b3+c3=1a,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)
(a,b,c)=(0,0,1)wecangetab+bc+ca=0,abc=0,thereforea,b,carerootsofx3−x2+0x+0=0⇒x2(x−1)=0⇒x∈(0,0,1)
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.
SirI′vetriedtosolveinadifferentwayinwhich2alsoappearsasaroot.Iknowthiscan′tsatisfythesystem.ButthenIcan′tdesidewhetherit′sextraneousrootbecauseIhaven′tsquaredanyequation.Pleaseseemysolutionandguidemeasyoualwaysdo.
Commented by Rasheed.Sindhi last updated on 18/Jul/21
Sorry sir I found my error when I solved it this time! Thanks Sir.
SorrysirIfoundmyerrorwhenIsolveditthistime!ThanksSir.
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.
ithinkyoualsogettherightanswerinthatway.yougetc=1thena+b=0andab=0,i.e.a=b=0.orc=0thena+b=1andab=0,i.e.a=1,b=0ora=0,b=1.
Commented by Rasheed.Sindhi last updated on 18/Jul/21
ThanX for this nice trick Sir, I didn′t think of that.
ThanXforthisnicetrickSir,Ididn′tthinkofthat.
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=1a2+b2+c2=1a3+b3+c3=1a2+b2+c2=(a+b+c)2−2(ab+bc+ca)1=1−2(ab+bc+ca)ab+bc+ca=0a3+b3+c3=(a+b+c)(a2+b2+c2)−(ab2+ac2+a2b+bc2+a2c+b2c)1=1×1−(ab2+ac2+a2b+bc2+a2c+b2c)0=a(ab+ca)+b(ab+bc)+c(bc+ca)0=a(−bc)+b(−ca)+c(−ab)abc=0weverifythata=0⇒b=0andc=1orb=1andc=0S={(1,0,0);(0,1,0);(0,0,1)}
Commented by Rasheed.Sindhi last updated on 18/Jul/21
Thanks Sir Olaf! I always appreciate your precious contribution towards this forum!
ThanksSirOlaf!Ialwaysappreciateyourpreciouscontributiontowardsthisforum!
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)
a+b+c=a2+b2+c2=a3+b3+c3=1−;a,b,c=?(i)⇒a+b=1−c(ii)⇒(a+b)2−2ab+c2−1=0(1−c)2−2ab+c2−1=0ab=1−2c+c2+c2−12=c2−c…A(iii)⇒(a+b)3−3ab(a+b)+c3−1=0(1−c)3−3ab(1−c)−(1−c3)=0(1−c){(1−c)2−3ab−(1+c+c2)}=0c=1∨\cancel1−2c+\cancelc2−3ab−\cancel1−c−\cancelc2=0ab=−c………………………BA&B:c2−c=−cc2=0⇒c=0Observingsymmetricnatureofthesystemwecanfixsamevaluesfora&b.∙∙∙(a,b,c)=(1,0,0),(0,1,0),(0,0,1)

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