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Question Number 184728 by mnjuly1970 last updated on 11/Jan/23
x^( 2) − 3x +1=0 α , β are roots : ( α^( 3) +(1/β) )^( 3) + ( β^^( 3) +(1/α) )^( 3) = ?
x2−3x+1=0α,βareroots:(α3+1β)3+(β3+1α)3=?
Answered by cortano1 last updated on 11/Jan/23
x^2 −3x+1=0 ; α and β are the roots find (α^3 +(1/β))^3 +(β^3 +(1/α))^3 . Solution. { ((α^3 −3α^2 +α=0)),((β^3 −3β^2 +β=0)) :}and { ((α+(1/α)=3)),((β+(1/β)=3)) :} { ((α^3 +(1/β)+α+β=3(1+α^2 ))),((β^3 +(1/α)+α+β=3(1+β^2 ))) :} ⇒ { ((α^3 +(1/β)=3α^2 )),((β^3 +(1/α)=3β^2 )) :} ⇒(α^3 +(1/β))^3 +(β^3 +(1/α))^3 = 27(α^6 +β^6 ) ⇒27{(α^3 +β^3 )^2 −2α^3 β^3 } ⇒27{18^2 −2.1}= 8694
x2−3x+1=0;αandβaretherootsfind(α3+1β)3+(β3+1α)3.Solution.{α3−3α2+α=0β3−3β2+β=0and{α+1α=3β+1β=3{α3+1β+α+β=3(1+α2)β3+1α+α+β=3(1+β2)⇒{α3+1β=3α2β3+1α=3β2⇒(α3+1β)3+(β3+1α)3=27(α6+β6)⇒27{(α3+β3)2−2α3β3}⇒27{182−2.1}=8694
Commented by mnjuly1970 last updated on 11/Jan/23
thx alot sir
thxalotsir
Answered by MJS_new last updated on 11/Jan/23
x^2 −3x+1=0 I′ll only use these 3 informations: (1) α+β=3 (2) αβ=1 [⇒ β=(1/α)] α^3 +(1/β)=α^3 +α=(α^2 /β)+α=((α(α+β))/β)=((3α)/β)=3α^2 same for β^3 +(1/α)=3β^2 now we have 27(α^6 +β^6 ) α^2 +β^2 =(α+β)^2 −2αβ=7 α^4 +β^4 =(α+β)^4 −2αβ(2(α^2 +β^2 )+3αβ)=47 α^6 +β^6 =(α+β)^6 −αβ(6(α^4 +β^4 )+15αβ(α^2 +β^2 )+20(αβ)^2 )=322 ⇒ 27(α^6 +β^6 )=8694
x2−3x+1=0I′llonlyusethese3informations:(1)α+β=3(2)αβ=1[⇒β=1α]α3+1β=α3+α=α2β+α=α(α+β)β=3αβ=3α2sameforβ3+1α=3β2nowwehave27(α6+β6)α2+β2=(α+β)2−2αβ=7α4+β4=(α+β)4−2αβ(2(α2+β2)+3αβ)=47α6+β6=(α+β)6−αβ(6(α4+β4)+15αβ(α2+β2)+20(αβ)2)=322⇒27(α6+β6)=8694
Answered by Rasheed.Sindhi last updated on 11/Jan/23
α+β=3 & αβ=1 α^2 −3α+1=0⇒α^3 =3α^2 −α α^( 3) +(1/β)=3α^2 −α+α=3α^2 Similarly, β^^( 3) +(1/α)=3β^2 ( α^( 3) +(1/β) )^( 3) + ( β^^( 3) +(1/α) )^( 3) =(3α^2 )^3 +(3β^2 )^3 27(α^6 +β^6 )=? α^3 +β^3 +3αβ(α+β)=27 α^3 +β^3 +3(1)(3)=27 α^3 +β^3 =18 (α^3 +β^3 )^2 =324 α^6 +β^6 +2(αβ)^3 =324 α^6 +β^6 =322 27(α^6 +β^6 )=27(322)=8694
α+β=3&αβ=1α2−3α+1=0⇒α3=3α2−αα3+1β=3α2−α+α=3α2Similarly,β3+1α=3β2(α3+1β)3+(β3+1α)3=(3α2)3+(3β2)327(α6+β6)=?α3+β3+3αβ(α+β)=27α3+β3+3(1)(3)=27α3+β3=18(α3+β3)2=324α6+β6+2(αβ)3=324α6+β6=32227(α6+β6)=27(322)=8694
Answered by Rasheed.Sindhi last updated on 11/Jan/23
α,β are roots of x^( 2) − 3x +1=0 (α^( 3) +(1/β) )^( 3) + (β^^( 3) +(1/α) )^( 3) = ? αβ=1 { ((α=1/β)),((β=1/α)) :} (α^( 3) +(1/β) )^( 3) + (β^( 3) +(1/α) )^( 3) (α^( 3) +α )^( 3) + (β^( 3) +β )^( 3) (α(α^2 +1) )^( 3) + (β(β^( 2) +1) )^( 3) x^( 2) − 3x +1=0⇒ { ((α^2 +1=3α)),((β^2 +1=3β)) :} (3α^2 )^3 +(3β^2 )^3 =27(α^6 +β^6 ) ▶α+β=3 α^3 +β^3 +3αβ(α+β)=27 α^3 +β^3 +3αβ(α+β)=27 α^3 +β^3 =18 (α^3 +β^3 )^2 =324 α^6 +β^6 +2α^3 β^3 =324 α^6 +β^6 =322 27(α^6 +β^6 )=27(322)=8694
α,βarerootsofx2−3x+1=0(α3+1β)3+(β3+1α)3=?αβ=1{α=1/ββ=1/α(α3+1β)3+(β3+1α)3(α3+α)3+(β3+β)3(α(α2+1))3+(β(β2+1))3x2−3x+1=0⇒{α2+1=3αβ2+1=3β(3α2)3+(3β2)3=27(α6+β6)▸α+β=3α3+β3+3αβ(α+β)=27α3+β3+3αβ(α+β)=27α3+β3=18(α3+β3)2=324α6+β6+2α3β3=324α6+β6=32227(α6+β6)=27(322)=8694
Answered by behi834171 last updated on 15/Jan/23
x^2 =3x−1⇒x^3 =3x^2 −x=3(3x−1)−x= =8x−3 x=3−(1/x)⇒(1/x)=3−x ⇒ { ((α^3 +β^3 =8(α+β)−6=24−6=18)),((α^2 +β^2 =3(α+β)−2=9−2=7)) :} ⇒(α^3 +(1/β))^3 =(8α−3+3−β)^3 =(8α−β)^3 = =[9α−(α+β)]^3 =[9α−3]^3 =3^3 [27α^3 −27α^2 +9α−1]= =27[27(8α−3)−27(3α−1)+9α−1]= =27[144α−55] ⇒Σ(𝛂^3 +(1/𝛃))^3 =27[144(α+β)−110]= =27[144×3−110]=27×322=8694 ■
x2=3x−1⇒x3=3x2−x=3(3x−1)−x==8x−3x=3−1x⇒1x=3−x⇒{α3+β3=8(α+β)−6=24−6=18α2+β2=3(α+β)−2=9−2=7⇒(α3+1β)3=(8α−3+3−β)3=(8α−β)3==[9α−(α+β)]3=[9α−3]3=33[27α3−27α2+9α−1]==27[27(8α−3)−27(3α−1)+9α−1]==27[144α−55]⇒Σ(α3+1β)3=27[144(α+β)−110]==27[144×3−110]=27×322=8694◼

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