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a-b-c-1-a-2-b-2-c-2-2-a-3-b-3-c-3-3-then-a-5-b-5-c-5-




Question Number 174514 by Best1 last updated on 03/Aug/22
a+b+c=1  a^2 +b^2 +c^2 =2  a^3 +b^3 +c^3 =3  then a^5 +b^5 +c^5  ?
a+b+c=1a2+b2+c2=2a3+b3+c3=3thena5+b5+c5?
Commented by mr W last updated on 03/Aug/22
see Q74970
seeQ74970
Answered by RedMath last updated on 03/Aug/22
no idea lol
noidealol
Commented by Best1 last updated on 03/Aug/22
sorry
sorry
Answered by MJS_new last updated on 03/Aug/22
1. a+b+c=α  2. a^2 +b^2 +c^2 =β  3.a^3 +b^3 +c^3 =γ  ========================  let a=u−(√v)∧b=u+(√v)  ⇒  1. c=α−2u  2. 6u^2 −4αu+2v+α^2 −β=0  3. −6u^3 +12αu^2 +6u(v−α^2 )+α^3 −γ=0  ========================  2. v=−3u^2 +2αu−((α^2 −β)/2)  3. u^3 −αu^2 +((3α^2 −β)/8)u=((α^3 −γ)/(24))  ========================  a^4 +b^4 +c^4 =  =−32α(u^3 −αu^2 +((3α^2 −β)/8)u)+((3α^4 −2α^2 β+β^2 )/2)=  =((α^4 −6α^2 β+8αγ+3β^2 )/6)  ========================  a^5 +b^5 +c^5 =  =−20(α^2 +β)(u^3 −αu^2 +((3α^2 −β)/8)u)+α^5 =  =((α^5 −5(α^3 β−α^2 γ−5lβγ))/6)  ========================
1.a+b+c=α2.a2+b2+c2=β3.a3+b3+c3=γ========================leta=uvb=u+v1.c=α2u2.6u24αu+2v+α2β=03.6u3+12αu2+6u(vα2)+α3γ=0========================2.v=3u2+2αuα2β23.u3αu2+3α2β8u=α3γ24========================a4+b4+c4==32α(u3αu2+3α2β8u)+3α42α2β+β22==α46α2β+8αγ+3β26========================a5+b5+c5==20(α2+β)(u3αu2+3α2β8u)+α5==α55(α3βα2γ5lβγ)6========================
Commented by Tawa11 last updated on 03/Aug/22
Great sirs.
Greatsirs.
Answered by behi834171 last updated on 03/Aug/22
a^2 +ab+ac=a  ab+b^2 +bc=b  ac+bc+c^2 =c⇒2Σab=Σa−Σa^2 =1−2=−1  ⇒Σab=−(1/2)  a^3 +a(b^2 +c^2 )=2a  b(a^2 +c^2 )+b^3 =2b  c(a^2 +b^2 )+c^3 =2c⇒  ⇒Σab(a+b)=2Σa−Σa^3 =2−3=−1  ⇒Σab(1−c)=−1⇒Σab−3abc=−1  ⇒−3abc=−1+(1/2)=−(1/2)⇒abc=(1/6)  so:a,b,c, are the roots of equation:       z^3 −z^2 −(1/2)z−(1/6)=0  z^5 =z^3 .z^2 =z^2 (z^2 +(1/2)z+(1/6))=z^4 +(1/2)z^3 +(1/6)z^2 =  =z(z^2 +(1/2)z+(1/6))+(1/2)(z^2 +(1/2)z+(1/6))+(1/6)z^2 =  =z^2 +(1/2)z+(1/6)+(1/2)z^2 +(1/6)z+(1/2)z^2 +(1/4)z+(1/(12))+(1/6)z^2 =  =((13)/6)z^2 +((11)/(12))z+(1/4)  ⇒Σa^5 =((13)/6)Σa^2 +((11)/(12))Σa+(3/4)=  =((13)/6)×2+((11)/(12))×1+(3/4)=6    .■  ......................................  by using symbols of dear mr:MJS_new  Σab=((Σa−Σa^2 )/2)=((𝛂−𝛃)/2)  Σab−3abc=2Σa−Σa^3 ⇒  abc=−(1/3)(2𝛂−𝛄−((𝛂−𝛃)/2))=−((3𝛂+𝛃−2𝛄)/6)  ⇒z^3 −𝛂z^2 +((𝛂−𝛃)/2)z+((3𝛂+𝛃−2𝛄)/6)=0  ⇒z^5 =z^2 z^3 =z^2 (𝛂z^2 −((𝛂−𝛃)/2)z−((3𝛂+𝛃−2𝛄)/6))=  =𝛂z(𝛂z^2 −((𝛂−𝛃)/2)z−((3𝛂+𝛃−2𝛄)/6))−((𝛂−𝛃)/2)(𝛂z^2 −((𝛂−𝛃)/2)z−((3𝛂+𝛃−2𝛄)/6))−((3𝛂+𝛃−2𝛄)/6)z^2 =  =𝛂^2 (𝛂z^2 −((𝛂−𝛃)/2)z−((3𝛂+𝛃−2𝛄)/6))−((𝛂(𝛂−𝛃))/2)z^2 −((𝛂(3𝛂+𝛃−2𝛄))/6)z−  +((𝛂(𝛂−𝛃))/2)z^2 +(((𝛂−𝛃)^2 )/4)z+(((𝛂−𝛃)(3𝛂+𝛃−2𝛄))/(12))−((3𝛂+𝛃−2𝛄)/6)z^2 =  =[𝛂^3 −((𝛂(𝛂−𝛃))/2)−((3𝛂+𝛃−2𝛄)/6)]z^2 −[((𝛂^2 (𝛂−𝛃))/2)+((𝛂(3𝛂+𝛃−2𝛄))/6)−(((𝛂−𝛃)^2 )/4)].z−  −(((2𝛂^2 −𝛂+𝛃)(3𝛂+𝛃−2𝛄))/(12))  =A.z^2 +B.z+C  A=((6𝛂^3 −3𝛂^2 +3𝛂𝛃−3𝛂−𝛃+2𝛄)/6)  B=−((6𝛂^3 −6𝛂^2 𝛃+3𝛂^2 +8𝛂𝛃−4𝛂𝛄−3𝛃^2 )/(12))  C=−((6𝛂^3 +2𝛂^2 𝛃−4𝛂^2 𝛄−3𝛂^2 +2𝛂𝛃+2𝛂𝛄+𝛃^2 −2𝛃𝛄)/(12))  ⇒Σa^5 =A.Σa^2 +B.Σa+3C
a2+ab+ac=aab+b2+bc=bac+bc+c2=c2Σab=ΣaΣa2=12=1Σab=12a3+a(b2+c2)=2ab(a2+c2)+b3=2bc(a2+b2)+c3=2cΣab(a+b)=2ΣaΣa3=23=1Σab(1c)=1Σab3abc=13abc=1+12=12abc=16so:a,b,c,aretherootsofequation:z3z212z16=0z5=z3.z2=z2(z2+12z+16)=z4+12z3+16z2==z(z2+12z+16)+12(z2+12z+16)+16z2==z2+12z+16+12z2+16z+12z2+14z+112+16z2==136z2+1112z+14Σa5=136Σa2+1112Σa+34==136×2+1112×1+34=6.◼..byusingsymbolsofdearmr:MJS_newΣab=ΣaΣa22=αβ2Σab3abc=2ΣaΣa3abc=13(2αγαβ2)=3α+β2γ6z3αz2+αβ2z+3α+β2γ6=0z5=z2z3=z2(αz2αβ2z3α+β2γ6)==αz(αz2αβ2z3α+β2γ6)αβ2(αz2αβ2z3α+β2γ6)3α+β2γ6z2==α2(αz2αβ2z3α+β2γ6)α(αβ)2z2α(3α+β2γ)6z+α(αβ)2z2+(αβ)24z+(αβ)(3α+β2γ)123α+β2γ6z2==[α3α(αβ)23α+β2γ6]z2[α2(αβ)2+α(3α+β2γ)6(αβ)24].z(2α2α+β)(3α+β2γ)12=A.z2+B.z+CA=6α33α2+3αβ3αβ+2γ6B=6α36α2β+3α2+8αβ4αγ3β212C=6α3+2α2β4α2γ3α2+2αβ+2αγ+β22βγ12Σa5=A.Σa2+B.Σa+3C
Commented by Best1 last updated on 03/Aug/22
  wawww thanks guys
wawwwthanksguys
Commented by Tawa11 last updated on 03/Aug/22
Great sirs
Greatsirs

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