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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) = ?
$$ \\ $$$$\:\:\:\:\:\:{x}^{\:\mathrm{2}} −\:\mathrm{3}{x}\:+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\alpha\:,\:\beta\:{are}\:{roots}\:: \\ $$$$\:\:\:\left(\:\alpha^{\:\mathrm{3}} \:+\frac{\mathrm{1}}{\beta}\:\right)^{\:\mathrm{3}} \:+\:\left(\:\beta^{\:^{\:\mathrm{3}} } \:+\frac{\mathrm{1}}{\alpha}\:\right)^{\:\mathrm{3}} =\:? \\ $$$$ \\ $$
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
$$\:{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}=\mathrm{0}\:;\:\alpha\:{and}\:\beta\:{are}\:{the}\:{roots} \\ $$$$\:{find}\:\left(\alpha^{\mathrm{3}} +\frac{\mathrm{1}}{\beta}\right)^{\mathrm{3}} +\left(\beta^{\mathrm{3}} +\frac{\mathrm{1}}{\alpha}\right)^{\mathrm{3}} . \\ $$$$\:{Solution}.\: \\ $$$$\:\begin{cases}{\alpha^{\mathrm{3}} −\mathrm{3}\alpha^{\mathrm{2}} +\alpha=\mathrm{0}}\\{\beta^{\mathrm{3}} −\mathrm{3}\beta^{\mathrm{2}} +\beta=\mathrm{0}}\end{cases}{and}\:\begin{cases}{\alpha+\frac{\mathrm{1}}{\alpha}=\mathrm{3}}\\{\beta+\frac{\mathrm{1}}{\beta}=\mathrm{3}}\end{cases} \\ $$$$\:\begin{cases}{\alpha^{\mathrm{3}} +\frac{\mathrm{1}}{\beta}+\alpha+\beta=\mathrm{3}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)}\\{\beta^{\mathrm{3}} +\frac{\mathrm{1}}{\alpha}+\alpha+\beta=\mathrm{3}\left(\mathrm{1}+\beta^{\mathrm{2}} \right)}\end{cases} \\ $$$$\:\Rightarrow\begin{cases}{\alpha^{\mathrm{3}} +\frac{\mathrm{1}}{\beta}=\mathrm{3}\alpha^{\mathrm{2}} }\\{\beta^{\mathrm{3}} +\frac{\mathrm{1}}{\alpha}=\mathrm{3}\beta^{\mathrm{2}} }\end{cases} \\ $$$$\Rightarrow\left(\alpha^{\mathrm{3}} +\frac{\mathrm{1}}{\beta}\right)^{\mathrm{3}} +\left(\beta^{\mathrm{3}} +\frac{\mathrm{1}}{\alpha}\right)^{\mathrm{3}} =\:\mathrm{27}\left(\alpha^{\mathrm{6}} +\beta^{\mathrm{6}} \right) \\ $$$$\Rightarrow\mathrm{27}\left\{\left(\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{2}\alpha^{\mathrm{3}} \beta^{\mathrm{3}} \right\} \\ $$$$\Rightarrow\mathrm{27}\left\{\mathrm{18}^{\mathrm{2}} −\mathrm{2}.\mathrm{1}\right\}=\:\mathrm{8694} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 11/Jan/23
thx alot sir
$$\:\:{thx}\:{alot}\:{sir} \\ $$
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
$${x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{I}'\mathrm{ll}\:\mathrm{only}\:\mathrm{use}\:\mathrm{these}\:\mathrm{3}\:\mathrm{informations}: \\ $$$$\left(\mathrm{1}\right)\:\alpha+\beta=\mathrm{3}\:\:\:\:\:\left(\mathrm{2}\right)\:\alpha\beta=\mathrm{1}\:\left[\Rightarrow\:\beta=\frac{\mathrm{1}}{\alpha}\right] \\ $$$$ \\ $$$$\alpha^{\mathrm{3}} +\frac{\mathrm{1}}{\beta}=\alpha^{\mathrm{3}} +\alpha=\frac{\alpha^{\mathrm{2}} }{\beta}+\alpha=\frac{\alpha\left(\alpha+\beta\right)}{\beta}=\frac{\mathrm{3}\alpha}{\beta}=\mathrm{3}\alpha^{\mathrm{2}} \\ $$$$\mathrm{same}\:\mathrm{for}\:\beta^{\mathrm{3}} +\frac{\mathrm{1}}{\alpha}=\mathrm{3}\beta^{\mathrm{2}} \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{27}\left(\alpha^{\mathrm{6}} +\beta^{\mathrm{6}} \right) \\ $$$$ \\ $$$$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta=\mathrm{7} \\ $$$$\alpha^{\mathrm{4}} +\beta^{\mathrm{4}} =\left(\alpha+\beta\right)^{\mathrm{4}} −\mathrm{2}\alpha\beta\left(\mathrm{2}\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \right)+\mathrm{3}\alpha\beta\right)=\mathrm{47} \\ $$$$\alpha^{\mathrm{6}} +\beta^{\mathrm{6}} =\left(\alpha+\beta\right)^{\mathrm{6}} −\alpha\beta\left(\mathrm{6}\left(\alpha^{\mathrm{4}} +\beta^{\mathrm{4}} \right)+\mathrm{15}\alpha\beta\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \right)+\mathrm{20}\left(\alpha\beta\right)^{\mathrm{2}} \right)=\mathrm{322} \\ $$$$\Rightarrow \\ $$$$\mathrm{27}\left(\alpha^{\mathrm{6}} +\beta^{\mathrm{6}} \right)=\mathrm{8694} \\ $$
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
$$\alpha+\beta=\mathrm{3}\:\&\:\alpha\beta=\mathrm{1} \\ $$$$\alpha^{\mathrm{2}} −\mathrm{3}\alpha+\mathrm{1}=\mathrm{0}\Rightarrow\alpha^{\mathrm{3}} =\mathrm{3}\alpha^{\mathrm{2}} −\alpha \\ $$$$\alpha^{\:\mathrm{3}} \:+\frac{\mathrm{1}}{\beta}=\mathrm{3}\alpha^{\mathrm{2}} −\alpha+\alpha=\mathrm{3}\alpha^{\mathrm{2}} \\ $$$${Similarly}, \\ $$$$\beta^{\:^{\:\mathrm{3}} } \:+\frac{\mathrm{1}}{\alpha}=\mathrm{3}\beta^{\mathrm{2}} \\ $$$$\left(\:\alpha^{\:\mathrm{3}} \:+\frac{\mathrm{1}}{\beta}\:\right)^{\:\mathrm{3}} \:+\:\left(\:\beta^{\:^{\:\mathrm{3}} } \:+\frac{\mathrm{1}}{\alpha}\:\right)^{\:\mathrm{3}} =\left(\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{3}} +\left(\mathrm{3}\beta^{\mathrm{2}} \right)^{\mathrm{3}} \\ $$$$\mathrm{27}\left(\alpha^{\mathrm{6}} +\beta^{\mathrm{6}} \right)=? \\ $$$$\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\mathrm{3}\alpha\beta\left(\alpha+\beta\right)=\mathrm{27} \\ $$$$\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\mathrm{3}\left(\mathrm{1}\right)\left(\mathrm{3}\right)=\mathrm{27} \\ $$$$\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} =\mathrm{18} \\ $$$$\left(\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} \right)^{\mathrm{2}} =\mathrm{324} \\ $$$$\alpha^{\mathrm{6}} +\beta^{\mathrm{6}} +\mathrm{2}\left(\alpha\beta\right)^{\mathrm{3}} =\mathrm{324} \\ $$$$\alpha^{\mathrm{6}} +\beta^{\mathrm{6}} =\mathrm{322} \\ $$$$\mathrm{27}\left(\alpha^{\mathrm{6}} +\beta^{\mathrm{6}} \right)=\mathrm{27}\left(\mathrm{322}\right)=\mathrm{8694} \\ $$
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
$$\:\alpha,\beta\:{are}\:{roots}\:{of}\:{x}^{\:\mathrm{2}} −\:\mathrm{3}{x}\:+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\left(\alpha^{\:\mathrm{3}} \:+\frac{\mathrm{1}}{\beta}\:\right)^{\:\mathrm{3}} \:+\:\left(\beta^{\:^{\:\mathrm{3}} } \:+\frac{\mathrm{1}}{\alpha}\:\right)^{\:\mathrm{3}} =\:?\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\alpha\beta=\mathrm{1\begin{cases}{\alpha=\mathrm{1}/\beta}\\{\beta=\mathrm{1}/\alpha}\end{cases}}\: \\ $$$$\:\left(\alpha^{\:\mathrm{3}} \:+\frac{\mathrm{1}}{\beta}\:\right)^{\:\mathrm{3}} \:+\:\left(\beta^{\:\mathrm{3}} \:+\frac{\mathrm{1}}{\alpha}\:\right)^{\:\mathrm{3}} \\ $$$$\:\left(\alpha^{\:\mathrm{3}} \:+\alpha\:\right)^{\:\mathrm{3}} \:+\:\left(\beta^{\:\mathrm{3}} \:+\beta\:\right)^{\:\mathrm{3}} \\ $$$$\:\left(\alpha\left(\alpha^{\mathrm{2}} +\mathrm{1}\right)\:\:\right)^{\:\mathrm{3}} \:+\:\left(\beta\left(\beta^{\:\mathrm{2}} \:+\mathrm{1}\right)\:\right)^{\:\mathrm{3}} \\ $$$$\:\:\:\:\:\:{x}^{\:\mathrm{2}} −\:\mathrm{3}{x}\:+\mathrm{1}=\mathrm{0}\Rightarrow\begin{cases}{\alpha^{\mathrm{2}} +\mathrm{1}=\mathrm{3}\alpha}\\{\beta^{\mathrm{2}} +\mathrm{1}=\mathrm{3}\beta}\end{cases}\: \\ $$$$\left(\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{3}} +\left(\mathrm{3}\beta^{\mathrm{2}} \right)^{\mathrm{3}} \\ $$$$=\mathrm{27}\left(\alpha^{\mathrm{6}} +\beta^{\mathrm{6}} \right) \\ $$$$ \\ $$$$\blacktriangleright\alpha+\beta=\mathrm{3} \\ $$$$\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\mathrm{3}\alpha\beta\left(\alpha+\beta\right)=\mathrm{27} \\ $$$$\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\mathrm{3}\alpha\beta\left(\alpha+\beta\right)=\mathrm{27} \\ $$$$\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} =\mathrm{18} \\ $$$$\left(\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} \right)^{\mathrm{2}} =\mathrm{324} \\ $$$$\alpha^{\mathrm{6}} +\beta^{\mathrm{6}} +\mathrm{2}\alpha^{\mathrm{3}} \beta^{\mathrm{3}} =\mathrm{324} \\ $$$$\alpha^{\mathrm{6}} +\beta^{\mathrm{6}} =\mathrm{322} \\ $$$$\mathrm{27}\left(\alpha^{\mathrm{6}} +\beta^{\mathrm{6}} \right)=\mathrm{27}\left(\mathrm{322}\right)=\mathrm{8694} \\ $$
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 ■
$${x}^{\mathrm{2}} =\mathrm{3}{x}−\mathrm{1}\Rightarrow{x}^{\mathrm{3}} =\mathrm{3}{x}^{\mathrm{2}} −{x}=\mathrm{3}\left(\mathrm{3}{x}−\mathrm{1}\right)−{x}= \\ $$$$=\mathrm{8}{x}−\mathrm{3} \\ $$$${x}=\mathrm{3}−\frac{\mathrm{1}}{{x}}\Rightarrow\frac{\mathrm{1}}{{x}}=\mathrm{3}−{x} \\ $$$$\Rightarrow\begin{cases}{\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} =\mathrm{8}\left(\alpha+\beta\right)−\mathrm{6}=\mathrm{24}−\mathrm{6}=\mathrm{18}}\\{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\mathrm{3}\left(\alpha+\beta\right)−\mathrm{2}=\mathrm{9}−\mathrm{2}=\mathrm{7}}\end{cases} \\ $$$$\Rightarrow\left(\alpha^{\mathrm{3}} +\frac{\mathrm{1}}{\beta}\right)^{\mathrm{3}} =\left(\mathrm{8}\alpha−\mathrm{3}+\mathrm{3}−\beta\right)^{\mathrm{3}} =\left(\mathrm{8}\alpha−\beta\right)^{\mathrm{3}} = \\ $$$$=\left[\mathrm{9}\alpha−\left(\alpha+\beta\right)\right]^{\mathrm{3}} =\left[\mathrm{9}\alpha−\mathrm{3}\right]^{\mathrm{3}} =\mathrm{3}^{\mathrm{3}} \left[\mathrm{27}\alpha^{\mathrm{3}} −\mathrm{27}\alpha^{\mathrm{2}} +\mathrm{9}\alpha−\mathrm{1}\right]= \\ $$$$=\mathrm{27}\left[\mathrm{27}\left(\mathrm{8}\alpha−\mathrm{3}\right)−\mathrm{27}\left(\mathrm{3}\alpha−\mathrm{1}\right)+\mathrm{9}\alpha−\mathrm{1}\right]= \\ $$$$=\mathrm{27}\left[\mathrm{144}\alpha−\mathrm{55}\right] \\ $$$$\Rightarrow\Sigma\left(\boldsymbol{\alpha}^{\mathrm{3}} +\frac{\mathrm{1}}{\boldsymbol{\beta}}\right)^{\mathrm{3}} =\mathrm{27}\left[\mathrm{144}\left(\alpha+\beta\right)−\mathrm{110}\right]= \\ $$$$=\mathrm{27}\left[\mathrm{144}×\mathrm{3}−\mathrm{110}\right]=\mathrm{27}×\mathrm{322}=\mathrm{8694}\:\blacksquare \\ $$

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