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Question Number 209187 by mnjuly1970 last updated on 03/Jul/24
:: α , β and γ are roots of the following equation . Find the value of ” F ” : Equation : x^( 3) −2x −1=0 F := α^( 5) + β^( 5) + γ^( 5)
$$ \\ $$$$\:\:\:::\:\:\:\alpha\:,\:\beta\:\:{and}\:\:\gamma\:\:{are}\:{roots}\:{of}\:{the} \\ $$$$\:\:\:\:\:{following}\:\:{equation}\:.\:{Find}\:{the} \\ $$$$\:\:\:\:\:{value}\:\:{of}\:\:\:''\:\:\mathrm{F}\:\:''\::\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{E}{quation}\::\:\:\:\:\:\:{x}^{\:\mathrm{3}} \:−\mathrm{2}{x}\:\:−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{F}\::=\:\alpha^{\:\mathrm{5}} \:+\:\beta^{\:\mathrm{5}} \:+\:\gamma^{\:\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$
Answered by A5T last updated on 03/Jul/24
α+β+γ=0; αβ+βγ+γα=−2;αβγ=1 x^3 =2x+1⇒x^5 =2x^3 +x^2 =2(2x+1)+x^2 =x^2 +4x+2 ⇒α^5 +β^5 +γ^5 =α^2 +β^2 +γ^2 +4(α+β+γ)+6 =(α+β+γ)^2 −2(αβ+βγ+γα)+4(0)+6=10
$$\alpha+\beta+\gamma=\mathrm{0};\:\alpha\beta+\beta\gamma+\gamma\alpha=−\mathrm{2};\alpha\beta\gamma=\mathrm{1} \\ $$$${x}^{\mathrm{3}} =\mathrm{2}{x}+\mathrm{1}\Rightarrow{x}^{\mathrm{5}} =\mathrm{2}{x}^{\mathrm{3}} +{x}^{\mathrm{2}} =\mathrm{2}\left(\mathrm{2}{x}+\mathrm{1}\right)+{x}^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2} \\ $$$$\Rightarrow\alpha^{\mathrm{5}} +\beta^{\mathrm{5}} +\gamma^{\mathrm{5}} =\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} +\mathrm{4}\left(\alpha+\beta+\gamma\right)+\mathrm{6} \\ $$$$=\left(\alpha+\beta+\gamma\right)^{\mathrm{2}} −\mathrm{2}\left(\alpha\beta+\beta\gamma+\gamma\alpha\right)+\mathrm{4}\left(\mathrm{0}\right)+\mathrm{6}=\mathrm{10} \\ $$
Commented by mnjuly1970 last updated on 03/Jul/24
thank you so much ...
$$\:{thank}\:{you}\:{so}\:{much}\:… \\ $$
Answered by lepuissantcedricjunior last updated on 05/Jul/24
x^3 −2x−1=(x+1)(x^2 −x−1) =(x+1)(x−((1−(√5))/2))(x−((1+(√5))/2)) 𝛂=−1;𝛃=((1−(√5))/2);𝛄=((1+(√5))/2) F:(−1)^5 +(((1−(√5))/2))^5 +(((1+(√5))/2))^5 =−1+(((1−(√5))/2))(((6−2(√5))/4))^2 +(((1+(√5))/2))(((6+2(√5))/4))^2 =−1+(((1−(√5))/2))(((56−24(√5))/(16)))+(((1+(√5))/2))(((56+24(√5))/(16))) =−1+(((56+120−80(√5))/(32)))+(((56+120+80(√5))/(32))) =−1+((176×2)/(32))=−1+((352)/(32)) =−1+11=10 =>F:𝛂^5 +𝛃^5 +𝛄^5 =10
$$\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{2}\boldsymbol{{x}}−\mathrm{1}=\left(\boldsymbol{{x}}+\mathrm{1}\right)\left(\boldsymbol{{x}}^{\mathrm{2}} −\boldsymbol{{x}}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\boldsymbol{{x}}+\mathrm{1}\right)\left(\boldsymbol{{x}}−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left(\boldsymbol{{x}}−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$$$\boldsymbol{\alpha}=−\mathrm{1};\boldsymbol{\beta}=\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}};\boldsymbol{\gamma}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\boldsymbol{{F}}:\left(−\mathrm{1}\right)^{\mathrm{5}} +\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{5}} +\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{5}} \\ $$$$\:\:\:\:=−\mathrm{1}+\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left(\frac{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left(\frac{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:=−\mathrm{1}+\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left(\frac{\mathrm{56}−\mathrm{24}\sqrt{\mathrm{5}}}{\mathrm{16}}\right)+\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left(\frac{\mathrm{56}+\mathrm{24}\sqrt{\mathrm{5}}}{\mathrm{16}}\right) \\ $$$$\:\:=−\mathrm{1}+\left(\frac{\mathrm{56}+\mathrm{120}−\mathrm{80}\sqrt{\mathrm{5}}}{\mathrm{32}}\right)+\left(\frac{\mathrm{56}+\mathrm{120}+\mathrm{80}\sqrt{\mathrm{5}}}{\mathrm{32}}\right) \\ $$$$\:\:=−\mathrm{1}+\frac{\mathrm{176}×\mathrm{2}}{\mathrm{32}}=−\mathrm{1}+\frac{\mathrm{352}}{\mathrm{32}} \\ $$$$\:\:=−\mathrm{1}+\mathrm{11}=\mathrm{10} \\ $$$$=>\boldsymbol{{F}}:\boldsymbol{\alpha}^{\mathrm{5}} +\boldsymbol{\beta}^{\mathrm{5}} +\boldsymbol{\gamma}^{\mathrm{5}} =\mathrm{10} \\ $$
Commented by Spillover last updated on 06/Jul/24
good
$${good} \\ $$
Answered by behi834171 last updated on 09/Jul/24
x^3 =2x+1⇒x^4 =2x^2 +x⇒x^5 =2x^3 +x^2 = 2(2x+1)+x^2 =x^2 +4x+2 α+β+γ=0⇒Σα^2 =(Σα)^2 −2(Σαβ)= =0−2(−2)=4 ⇒𝚺α^5 =𝚺𝛂^2 +4𝚺𝛂+3=4+4(0)+6=10.■
$${x}^{\mathrm{3}} =\mathrm{2}{x}+\mathrm{1}\Rightarrow{x}^{\mathrm{4}} =\mathrm{2}{x}^{\mathrm{2}} +{x}\Rightarrow{x}^{\mathrm{5}} =\mathrm{2}{x}^{\mathrm{3}} +{x}^{\mathrm{2}} = \\ $$$$\mathrm{2}\left(\mathrm{2}{x}+\mathrm{1}\right)+{x}^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2} \\ $$$$\alpha+\beta+\gamma=\mathrm{0}\Rightarrow\Sigma\alpha^{\mathrm{2}} =\left(\Sigma\alpha\right)^{\mathrm{2}} −\mathrm{2}\left(\Sigma\alpha\beta\right)= \\ $$$$=\mathrm{0}−\mathrm{2}\left(−\mathrm{2}\right)=\mathrm{4} \\ $$$$\Rightarrow\boldsymbol{\Sigma}\alpha^{\mathrm{5}} =\boldsymbol{\Sigma\alpha}^{\mathrm{2}} +\mathrm{4}\boldsymbol{\Sigma\alpha}+\mathrm{3}=\mathrm{4}+\mathrm{4}\left(\mathrm{0}\right)+\mathrm{6}=\mathrm{10}.\blacksquare \\ $$

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