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Question Number 219454 by mr W last updated on 25/Apr/25
a, b, c are the roots of the equation x^3 −3x+1=0. find (a)^(1/3) +(b)^(1/3) +(c)^(1/3) =? & (1/( (a)^(1/3) ))+(1/( (b)^(1/3) ))+(1/( (c)^(1/3) ))=?
$${a},\:{b},\:{c}\:{are}\:{the}\:{roots}\:{of}\:{the}\:{equation} \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{1}=\mathrm{0}. \\ $$$${find}\:\sqrt[{\mathrm{3}}]{{a}}+\sqrt[{\mathrm{3}}]{{b}}+\sqrt[{\mathrm{3}}]{{c}}=?\:\&\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{c}}}=? \\ $$
Commented by mr W last updated on 25/Apr/25
(see also old question 217985)
$$\left({see}\:{also}\:{old}\:{question}\:\mathrm{217985}\right) \\ $$
Commented by Ghisom last updated on 26/Apr/25
I found these: a^(1/3) +b^(1/3) +c^(1/3) =(3^(5/3) −6)^(1/3) a^(−1/3) +b^(−1/3) +c^(−1/3) =(3^(3/5) −3)^(1/3)
$$\mathrm{I}\:\mathrm{found}\:\mathrm{these}: \\ $$$${a}^{\mathrm{1}/\mathrm{3}} +{b}^{\mathrm{1}/\mathrm{3}} +{c}^{\mathrm{1}/\mathrm{3}} =\left(\mathrm{3}^{\mathrm{5}/\mathrm{3}} −\mathrm{6}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${a}^{−\mathrm{1}/\mathrm{3}} +{b}^{−\mathrm{1}/\mathrm{3}} +{c}^{−\mathrm{1}/\mathrm{3}} =\left(\mathrm{3}^{\mathrm{3}/\mathrm{5}} −\mathrm{3}\right)^{\mathrm{1}/\mathrm{3}} \\ $$
Commented by mr W last updated on 27/Apr/25
correct! thanks!
$${correct}!\:{thanks}! \\ $$
Answered by mr W last updated on 27/Apr/25
this is how i solved: a, b, c are roots of x^3 −3x+1=0 thus: a+b+c=0 abc=−1 (1/x^3 )−(3/x^2 )+1=0 thus: (1/a)+(1/b)+(1/c)=3 say u=(a)^(1/3) +(b)^(1/3) +(c)^(1/3) , v=(1/( (a)^(1/3) ))+(1/( (b)^(1/3) ))+(1/( (c)^(1/3) )) u^3 =((a)^(1/3) )^3 +((b)^(1/3) )^3 +((c)^(1/3) )^3 −3((abc))^(1/3) +3((a)^(1/3) +(b)^(1/3) +(c)^(1/3) )(((ab))^(1/3) +((bc))^(1/3) +((ca))^(1/3) ) u^3 =a+b+c−3((abc))^(1/3) +3((a)^(1/3) +(b)^(1/3) +(c)^(1/3) )((abc))^(1/3) ((1/( (a)^(1/3) ))+(1/( (b)^(1/3) ))+(1/( (c)^(1/3) ))) ⇒u^3 =3−3uv ...(i) v^3 =((1/( (a)^(1/3) )))^3 +((1/( (b)^(1/3) )))^3 +((1/( (c)^(1/3) )))^3 −(3/( ((abc))^(1/3) ))+3((1/( (a)^(1/3) ))+(1/( (b)^(1/3) ))+(1/( (c)^(1/3) )))((1/( ((ab))^(1/3) ))+(1/( ((bc))^(1/3) ))+(1/( ((ca))^(1/3) ))) v^3 =(1/a)+(1/b)+(1/c)−(3/( ((abc))^(1/3) ))+3((1/( (a)^(1/3) ))+(1/( (b)^(1/3) ))+(1/( (c)^(1/3) )))(1/( ((abc))^(1/3) ))((a)^(1/3) +(b)^(1/3) +(c)^(1/3) ) ⇒v^3 =6−3vu ...(ii) (i)×(ii): (uv)^3 =18−27(uv)+9(uv)^2 (uv)^3 −9(uv)^2 +27(uv)−18=0 let p=uv p^3 −9p^2 +27p−18=0 let p=s+3 (s+3)^3 −9(s+3)^2 +27(s+3)−18=0 s^3 +9s^2 +27s+27−9s^2 −54s−81+27s+81−18=0 s^3 +9=0 ⇒s=−(9)^(1/3) ⇒uv=p=3+s=3−(9)^(1/3) u^3 =3(1−uv)=3((9)^(1/3) −2) ⇒u=((3((9)^(1/3) −2)))^(1/3) ✓ v^3 =3(2−uv)=3((9)^(1/3) −1) ⇒v=((3((9)^(1/3) −1)))^(1/3) ✓
$${this}\:{is}\:{how}\:{i}\:{solved}: \\ $$$$ \\ $$$${a},\:{b},\:{c}\:{are}\:{roots}\:{of} \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{1}=\mathrm{0} \\ $$$${thus}: \\ $$$${a}+{b}+{c}=\mathrm{0} \\ $$$${abc}=−\mathrm{1} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{3}} }−\frac{\mathrm{3}}{{x}^{\mathrm{2}} }+\mathrm{1}=\mathrm{0} \\ $$$${thus}: \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\mathrm{3} \\ $$$$ \\ $$$${say}\:{u}=\sqrt[{\mathrm{3}}]{{a}}+\sqrt[{\mathrm{3}}]{{b}}+\sqrt[{\mathrm{3}}]{{c}},\: \\ $$$$\:\:\:\:\:\:\:\:{v}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{c}}} \\ $$$${u}^{\mathrm{3}} =\left(\sqrt[{\mathrm{3}}]{{a}}\right)^{\mathrm{3}} +\left(\sqrt[{\mathrm{3}}]{{b}}\right)^{\mathrm{3}} +\left(\sqrt[{\mathrm{3}}]{{c}}\right)^{\mathrm{3}} −\mathrm{3}\sqrt[{\mathrm{3}}]{{abc}}+\mathrm{3}\left(\sqrt[{\mathrm{3}}]{{a}}+\sqrt[{\mathrm{3}}]{{b}}+\sqrt[{\mathrm{3}}]{{c}}\right)\left(\sqrt[{\mathrm{3}}]{{ab}}+\sqrt[{\mathrm{3}}]{{bc}}+\sqrt[{\mathrm{3}}]{{ca}}\right) \\ $$$${u}^{\mathrm{3}} ={a}+{b}+{c}−\mathrm{3}\sqrt[{\mathrm{3}}]{{abc}}+\mathrm{3}\left(\sqrt[{\mathrm{3}}]{{a}}+\sqrt[{\mathrm{3}}]{{b}}+\sqrt[{\mathrm{3}}]{{c}}\right)\sqrt[{\mathrm{3}}]{{abc}}\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{c}}}\right) \\ $$$$\Rightarrow{u}^{\mathrm{3}} =\mathrm{3}−\mathrm{3}{uv}\:\:\:…\left({i}\right) \\ $$$$ \\ $$$${v}^{\mathrm{3}} =\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}\right)^{\mathrm{3}} +\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}\right)^{\mathrm{3}} +\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{c}}}\right)^{\mathrm{3}} −\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}}]{{abc}}}+\mathrm{3}\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{c}}}\right)\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{ab}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{bc}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{ca}}}\right) \\ $$$${v}^{\mathrm{3}} =\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}−\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}}]{{abc}}}+\mathrm{3}\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{c}}}\right)\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{abc}}}\left(\sqrt[{\mathrm{3}}]{{a}}+\sqrt[{\mathrm{3}}]{{b}}+\sqrt[{\mathrm{3}}]{{c}}\right) \\ $$$$\Rightarrow{v}^{\mathrm{3}} =\mathrm{6}−\mathrm{3}{vu}\:\:\:…\left({ii}\right) \\ $$$$ \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$$\left({uv}\right)^{\mathrm{3}} =\mathrm{18}−\mathrm{27}\left({uv}\right)+\mathrm{9}\left({uv}\right)^{\mathrm{2}} \\ $$$$\left({uv}\right)^{\mathrm{3}} −\mathrm{9}\left({uv}\right)^{\mathrm{2}} +\mathrm{27}\left({uv}\right)−\mathrm{18}=\mathrm{0} \\ $$$${let}\:{p}={uv} \\ $$$${p}^{\mathrm{3}} −\mathrm{9}{p}^{\mathrm{2}} +\mathrm{27}{p}−\mathrm{18}=\mathrm{0} \\ $$$${let}\:{p}={s}+\mathrm{3} \\ $$$$\left({s}+\mathrm{3}\right)^{\mathrm{3}} −\mathrm{9}\left({s}+\mathrm{3}\right)^{\mathrm{2}} +\mathrm{27}\left({s}+\mathrm{3}\right)−\mathrm{18}=\mathrm{0} \\ $$$${s}^{\mathrm{3}} +\mathrm{9}{s}^{\mathrm{2}} +\mathrm{27}{s}+\mathrm{27}−\mathrm{9}{s}^{\mathrm{2}} −\mathrm{54}{s}−\mathrm{81}+\mathrm{27}{s}+\mathrm{81}−\mathrm{18}=\mathrm{0} \\ $$$${s}^{\mathrm{3}} +\mathrm{9}=\mathrm{0} \\ $$$$\Rightarrow{s}=−\sqrt[{\mathrm{3}}]{\mathrm{9}} \\ $$$$\Rightarrow{uv}={p}=\mathrm{3}+{s}=\mathrm{3}−\sqrt[{\mathrm{3}}]{\mathrm{9}} \\ $$$$ \\ $$$${u}^{\mathrm{3}} =\mathrm{3}\left(\mathrm{1}−{uv}\right)=\mathrm{3}\left(\sqrt[{\mathrm{3}}]{\mathrm{9}}−\mathrm{2}\right) \\ $$$$\Rightarrow{u}=\sqrt[{\mathrm{3}}]{\mathrm{3}\left(\sqrt[{\mathrm{3}}]{\mathrm{9}}−\mathrm{2}\right)}\:\:\checkmark \\ $$$${v}^{\mathrm{3}} =\mathrm{3}\left(\mathrm{2}−{uv}\right)=\mathrm{3}\left(\sqrt[{\mathrm{3}}]{\mathrm{9}}−\mathrm{1}\right) \\ $$$$\Rightarrow{v}=\sqrt[{\mathrm{3}}]{\mathrm{3}\left(\sqrt[{\mathrm{3}}]{\mathrm{9}}−\mathrm{1}\right)}\:\:\checkmark \\ $$

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