f-x-x-3-16x-2-57x-1-f-a-0-f-b-0-f-c-0-a-1-5-b-1-5-c-1-5-

Question Number 204595 by SEKRET last updated on 22/Feb/24

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\:=\:\boldsymbol{\mathrm{x}}^{\mathrm{3}} \:−\:\mathrm{16}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:−\:\mathrm{57}\boldsymbol{\mathrm{x}}\:+\mathrm{1}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{a}}\right)=\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{b}}\right)=\mathrm{0}\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{c}}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\sqrt[{\mathrm{5}}]{\boldsymbol{\mathrm{a}}}\:+\:\sqrt[{\mathrm{5}}]{\boldsymbol{\mathrm{b}}\:}\:+\:\sqrt[{\mathrm{5}}]{\boldsymbol{\mathrm{c}}\:}\:=\:? \\ $$

Answered by mr W last updated on 22/Feb/24

t=(x)^(1/5) ⇒x=t^5 t^(15) −16t^(10) −57t^5 +1=0 (t^3 −t^2 −2t+1)(t^(12) +t^(11) +...+2t+1)=0 t^3 −t^2 −2t+1=0 ⇒(a)^(1/5) +(b)^(1/5) +(c)^(1/5) =t_1 +t_2 +t_3 =1 ✓

$${t}=\sqrt[{\mathrm{5}}]{{x}}\Rightarrow{x}={t}^{\mathrm{5}} \\ $$$${t}^{\mathrm{15}} −\mathrm{16}{t}^{\mathrm{10}} −\mathrm{57}{t}^{\mathrm{5}} +\mathrm{1}=\mathrm{0} \\ $$$$\left({t}^{\mathrm{3}} −{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}\right)\left({t}^{\mathrm{12}} +{t}^{\mathrm{11}} +…+\mathrm{2}{t}+\mathrm{1}\right)=\mathrm{0} \\ $$$${t}^{\mathrm{3}} −{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\sqrt[{\mathrm{5}}]{{a}}+\sqrt[{\mathrm{5}}]{{b}}+\sqrt[{\mathrm{5}}]{{c}}={t}_{\mathrm{1}} +{t}_{\mathrm{2}} +{t}_{\mathrm{3}} =\mathrm{1}\:\checkmark \\ $$

Commented by SEKRET last updated on 22/Feb/24