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Question Number 190520 by mathlove last updated on 04/Apr/23
if a,b and c root of the x^3 −16x^2 −57x+1=0 thi find thd volue of a^(1/5) +b^(1/5) +c^(1/5) =?
ifa,bandcrootofthex3−16x2−57x+1=0thifindthdvolueofa15+b15+c15=?
Answered by Frix last updated on 05/Apr/23
If the solution is ∈Z we need to find a polynomial factor of degree 3 of t^(15) −16t^(10) −57t^5 +1 I used software to find the factor t^3 −1t^2 −2t+1 The other factor of degree 12 is prime ⇒ answer is 1 [I don′t know if there′s an analytical method to solve this]
Ifthesolutionis∈Zweneedtofindapolynomialfactorofdegree3oft15−16t10−57t5+1Iusedsoftwaretofindthefactort3−1t2−2t+1Theotherfactorofdegree12isprime⇒answeris1[Idon′tknowifthere′sananalyticalmethodtosolvethis]
Answered by behi834171 last updated on 05/Apr/23
x^3 =16x^2 +57x−1 x^4 =16x^3 +57x^2 −x=16(16x^2 +57x−1)+57x^2 −x= =313x^2 +911x−16 x^5 =313x^3 +911x^2 −16x=313(16x^2 +57x−1)+911x^2 −16x= =5919x^2 +17852x−313 ⇒x^5 −5919x^2 −17852x+313=0 y=(1/x)⇒(1/y^5 )−((5919)/y^2 )−((17852)/y)+313=0 ⇒313y^5 −17852y^4 −5919y^3 +1=0 ⇒Σa^(1/5) =−((−17852)/(313))=((17852)/(313)) .■
x3=16x2+57x−1x4=16x3+57x2−x=16(16x2+57x−1)+57x2−x==313x2+911x−16x5=313x3+911x2−16x=313(16x2+57x−1)+911x2−16x==5919x2+17852x−313⇒\boldsymbolx5−5919\boldsymbolx2−17852\boldsymbolx+313=0y=1x⇒1y5−5919y2−17852y+313=0⇒313\boldsymboly5−17852\boldsymboly4−5919\boldsymboly3+1=0⇒Σ\boldsymbola15=−−17852313=17852313.◼
Commented by mr W last updated on 07/Apr/23
thanks sir! your method seems to be a good idea, but i think it is not correct. when you transform the equation to 313y^5 −17852y^4 −5919y^3 +1=0, it has 5 roots. so you get with Σy=Σ(1/x)=((17852)/(313)) the sum of all its 5 roots, i.e. (1/a)+(1/b)+(1/c)+(1/d)+(1/e)=((17852)/(313)). but this is not that what the original question has requested: a^(1/5) +b^(1/5) +c^(1/5) =?
thankssir!yourmethodseemstobeagoodidea,butithinkitisnotcorrect.whenyoutransformtheequationto313\boldsymboly5−17852\boldsymboly4−5919\boldsymboly3+1=0,ithas5roots.soyougetwithΣy=Σ1x=17852313thesumofallits5roots,i.e.1a+1b+1c+1d+1e=17852313.butthisisnotthatwhattheoriginalquestionhasrequested:a15+b15+c15=?
Commented by behi834171 last updated on 08/Apr/23
hello my dear master. you are right, and ,i have a terrible typo i dont delete my way,but will correct it and i will post that. thanks for pointing this.
hellomydearmaster.youareright,and,ihaveaterribletypoidontdeletemyway,butwillcorrectitandiwillpostthat.thanksforpointingthis.

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