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Question-223703

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Question Number 223703 by Rojarani last updated on 02/Aug/25
Answered by Rasheed.Sindhi last updated on 02/Aug/25
(((x−1)/3))^3 =(4/9)+((3/(27)))^(1/4) +((9/(27)))^(1/3) ((x^3 −3x^2 +3x−1)/(27))=(4/9)+((3)^(1/3) /3) +((9)^(1/3) /3) x^3 −3x^2 +3x−1=12+9(3)^(1/3) +9(9)^(1/3) =12+3(3(3)^(1/3) ) +(3(3)^(1/3) )^2 =12+3a+a^2 ; a=3(3)^(1/3) x^3 =3x^2 −3x+1+a^2 +3a+12 (1/(3x^3 ))+(1/(2x^2 ))+(1/(4x))=((4+6x+3x^2 )/(12x^3 )) =((4+6x+3x^2 )/(12(3x^2 −3x+1+a^2 +3a+12))) ....
$$\left(\frac{{x}−\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} =\frac{\mathrm{4}}{\mathrm{9}}+\sqrt[{\mathrm{4}}]{\frac{\mathrm{3}}{\mathrm{27}}}\:+\sqrt[{\mathrm{3}}]{\frac{\mathrm{9}}{\mathrm{27}}}\: \\ $$$$\frac{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{1}}{\mathrm{27}}=\frac{\mathrm{4}}{\mathrm{9}}+\frac{\sqrt[{\mathrm{3}}]{\mathrm{3}}}{\mathrm{3}}\:+\frac{\sqrt[{\mathrm{3}}]{\mathrm{9}}}{\mathrm{3}}\: \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{1}=\mathrm{12}+\mathrm{9}\sqrt[{\mathrm{3}}]{\mathrm{3}}\:+\mathrm{9}\sqrt[{\mathrm{3}}]{\mathrm{9}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{12}+\mathrm{3}\left(\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{3}}\:\right)\:+\left(\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{3}}\:\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{12}+\mathrm{3}{a}+{a}^{\mathrm{2}} \:;\:\:{a}=\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{3}}\: \\ $$$${x}^{\mathrm{3}} =\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}+{a}^{\mathrm{2}} +\mathrm{3}{a}+\mathrm{12} \\ $$$$\: \\ $$$$\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}{x}}=\frac{\mathrm{4}+\mathrm{6}{x}+\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{12}{x}^{\mathrm{3}} } \\ $$$$\:\:\:\:\:=\frac{\mathrm{4}+\mathrm{6}{x}+\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{12}\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}+{a}^{\mathrm{2}} +\mathrm{3}{a}+\mathrm{12}\right)} \\ $$$$…. \\ $$
Answered by Wuji last updated on 02/Aug/25
^3 (√((64)/(729))) =((4^3 /9^3 ))^(1/3) =((4/9))^(3×(1/3)) =(4/9) ^3 (√(1/9)) =((1/9))^(1/3) =(1/((3^2 )^(1/3) )) =(1/3^(2/3) ) :^3 (√(1/3)) =(1/3^(1/3) ) ⇒(((x−1)/3))^3 =(4/9)+(1/3^(2/3) )+(1/3^(1/3) ) set k=(1/3^(1/3) ) ⇒k^3 =(1/3) ⇒k^2 =(1/3^(2/3) ) ⇒^3 (√((64)/(729)))+^3 (√(1/9))+^3 (√(1/3))=(4/9)×k^2 +k from (k^2 +k)^3 =k^6 +3k^5 +3k^4 +k^3 k^6 =(k^3 )^2 =((1/3))^2 =(1/9) ; k^5 =k^2 •k^3 =(k^2 /3); k^4 =(k/3);k^3 =(1/3) ⇒(k^2 +k)^3 =(1/9)+3((k^2 /3))+3((k/3))+(1/3) (k^2 +k)^3 =(1/9)+k^2 +k+(1/3)=(4/9)+k^2 +k ⇒^3 (√((64)/(729)))+^3 (√(1/9))+^3 (√(1/3))=(4/9)×k^2 +k ⇒(((x−1)/3))^3 =(k^2 +k)^3 ⇒((x−1)/3)=(k^2 +k) ⇒x−1=3(k^2 +k) x=1+3(k^2 +k) ⇒x=1+3^(1/3) +3^(2/3) m=3^(1/3) ⇒m^3 =3 ⇒x=1+m+m^2 (1/(3x^3 ))+(1/(2x^2 ))+(1/(4x)) =(4/(12x^3 ))+((6x)/(12x^3 ))+((3x^2 )/(12x^3 ))=((4+6x+3x^2 )/(12x^3 )) from x=1+m+m^3 (1+m+m^2 )−x=0 ⇒m+m^2 +1−x=0 m(m+m^2 +1−x)=0 ⇒m^2 +m^3 +m(1−x)=0 m^2 +3+m(1−x)=0 ⇒m^2 +m+3+m−mx=0 m^2 +2m+3=mx ⇒mx=x+2 ⇒m=((x+2)/x) ⇒x+1+m+m^2 =1+((x+2)/x)+(((x+2)^2 )/x^2 ) =((x^2 +x(x+2)+(x+2)^2 )/x^2 ) x=((x^2 +x^2 +2x+x^2 +4x+4)/x^2 ) ⇒x^3 =3x^2 +6x+4 ⇒(1/(3x^3 ))+(1/(2x^2 ))+(1/(4x))=(x^3 /(12x^3 ))=(1/(12)) A nice problem from my Boss Natai Chandra Bhar
$$ \\ $$$$\:^{\mathrm{3}} \sqrt{\frac{\mathrm{64}}{\mathrm{729}}}\:=\left(\frac{\mathrm{4}^{\mathrm{3}} }{\mathrm{9}^{\mathrm{3}} }\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:=\left(\frac{\mathrm{4}}{\mathrm{9}}\right)^{\mathrm{3}×\frac{\mathrm{1}}{\mathrm{3}}} =\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$\:^{\mathrm{3}} \sqrt{\frac{\mathrm{1}}{\mathrm{9}}}\:=\left(\frac{\mathrm{1}}{\mathrm{9}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\frac{\mathrm{1}}{\left(\mathrm{3}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{3}} }\:=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}/\mathrm{3}} }\:\::\:^{\mathrm{3}} \sqrt{\frac{\mathrm{1}}{\mathrm{3}}}\:=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{1}/\mathrm{3}} } \\ $$$$\Rightarrow\left(\frac{{x}−\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} =\frac{\mathrm{4}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}/\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{1}/\mathrm{3}} }\: \\ $$$${set}\:{k}=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{1}/\mathrm{3}} }\:\Rightarrow{k}^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{3}}\:\:\Rightarrow{k}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}/\mathrm{3}} } \\ $$$$\Rightarrow^{\mathrm{3}} \sqrt{\frac{\mathrm{64}}{\mathrm{729}}}+^{\mathrm{3}} \sqrt{\frac{\mathrm{1}}{\mathrm{9}}}+^{\mathrm{3}} \sqrt{\frac{\mathrm{1}}{\mathrm{3}}}=\frac{\mathrm{4}}{\mathrm{9}}×{k}^{\mathrm{2}} +{k} \\ $$$${from}\:\left({k}^{\mathrm{2}} +{k}\right)^{\mathrm{3}} ={k}^{\mathrm{6}} +\mathrm{3}{k}^{\mathrm{5}} +\mathrm{3}{k}^{\mathrm{4}} +{k}^{\mathrm{3}} \\ $$$${k}^{\mathrm{6}} =\left({k}^{\mathrm{3}} \right)^{\mathrm{2}} =\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{9}}\:;\:{k}^{\mathrm{5}} ={k}^{\mathrm{2}} \bullet{k}^{\mathrm{3}} =\frac{{k}^{\mathrm{2}} }{\mathrm{3}};\:{k}^{\mathrm{4}} =\frac{{k}}{\mathrm{3}};{k}^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\left({k}^{\mathrm{2}} +{k}\right)^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{9}}+\mathrm{3}\left(\frac{{k}^{\mathrm{2}} }{\mathrm{3}}\right)+\mathrm{3}\left(\frac{{k}}{\mathrm{3}}\right)+\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\left({k}^{\mathrm{2}} +{k}\right)^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{9}}+{k}^{\mathrm{2}} +{k}+\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{4}}{\mathrm{9}}+{k}^{\mathrm{2}} +{k} \\ $$$$\Rightarrow^{\mathrm{3}} \sqrt{\frac{\mathrm{64}}{\mathrm{729}}}+^{\mathrm{3}} \sqrt{\frac{\mathrm{1}}{\mathrm{9}}}+^{\mathrm{3}} \sqrt{\frac{\mathrm{1}}{\mathrm{3}}}=\frac{\mathrm{4}}{\mathrm{9}}×{k}^{\mathrm{2}} +{k} \\ $$$$\Rightarrow\left(\frac{{x}−\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} \:=\left({k}^{\mathrm{2}} +{k}\right)^{\mathrm{3}} \:\Rightarrow\frac{{x}−\mathrm{1}}{\mathrm{3}}=\left({k}^{\mathrm{2}} +{k}\right)\:\Rightarrow{x}−\mathrm{1}=\mathrm{3}\left({k}^{\mathrm{2}} +{k}\right) \\ $$$${x}=\mathrm{1}+\mathrm{3}\left({k}^{\mathrm{2}} +{k}\right)\:\Rightarrow{x}=\mathrm{1}+\mathrm{3}^{\mathrm{1}/\mathrm{3}} +\mathrm{3}^{\mathrm{2}/\mathrm{3}} \\ $$$${m}=\mathrm{3}^{\mathrm{1}/\mathrm{3}} \:\Rightarrow{m}^{\mathrm{3}} =\mathrm{3}\:\:\Rightarrow{x}=\mathrm{1}+{m}+{m}^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}{x}}\:=\frac{\mathrm{4}}{\mathrm{12}{x}^{\mathrm{3}} }+\frac{\mathrm{6}{x}}{\mathrm{12}{x}^{\mathrm{3}} }+\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{12}{x}^{\mathrm{3}} }=\frac{\mathrm{4}+\mathrm{6}{x}+\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{12}{x}^{\mathrm{3}} } \\ $$$${from}\:{x}=\mathrm{1}+{m}+{m}^{\mathrm{3}} \\ $$$$\left(\mathrm{1}+{m}+{m}^{\mathrm{2}} \right)−{x}=\mathrm{0}\:\Rightarrow{m}+{m}^{\mathrm{2}} +\mathrm{1}−{x}=\mathrm{0} \\ $$$${m}\left({m}+{m}^{\mathrm{2}} +\mathrm{1}−{x}\right)=\mathrm{0}\:\Rightarrow{m}^{\mathrm{2}} +{m}^{\mathrm{3}} +{m}\left(\mathrm{1}−{x}\right)=\mathrm{0} \\ $$$${m}^{\mathrm{2}} +\mathrm{3}+{m}\left(\mathrm{1}−{x}\right)=\mathrm{0}\:\Rightarrow{m}^{\mathrm{2}} +{m}+\mathrm{3}+{m}−{mx}=\mathrm{0} \\ $$$${m}^{\mathrm{2}} +\mathrm{2}{m}+\mathrm{3}={mx}\:\Rightarrow{mx}={x}+\mathrm{2}\:\Rightarrow{m}=\frac{{x}+\mathrm{2}}{{x}} \\ $$$$\Rightarrow{x}+\mathrm{1}+{m}+{m}^{\mathrm{2}} =\mathrm{1}+\frac{{x}+\mathrm{2}}{{x}}+\frac{\left({x}+\mathrm{2}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} }\:=\frac{{x}^{\mathrm{2}} +{x}\left({x}+\mathrm{2}\right)+\left({x}+\mathrm{2}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} } \\ $$$${x}=\frac{{x}^{\mathrm{2}} +{x}^{\mathrm{2}} +\mathrm{2}{x}+{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}}{{x}^{\mathrm{2}} }\:\Rightarrow{x}^{\mathrm{3}} =\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{4} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}{x}}=\frac{{x}^{\mathrm{3}} }{\mathrm{12}{x}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\boldsymbol{{A}}\:\boldsymbol{{nice}}\:\boldsymbol{{problem}}\:\boldsymbol{{from}}\:\boldsymbol{{my}}\:\boldsymbol{{Boss}}\:\boldsymbol{{Natai}}\:\boldsymbol{{Chandra}}\:\boldsymbol{{Bhar}} \\ $$

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