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Question-223403

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Question Number 223403 by Rojarani last updated on 24/Jul/25
Commented by mr W last updated on 24/Jul/25
should be ...=+ ((3((37))^(1/3) −10))^(1/3)
$${should}\:{be}\:\:…=+\:\sqrt[{\mathrm{3}}]{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{37}}−\mathrm{10}} \\ $$
Answered by mr W last updated on 25/Jul/25
once i developed following method while solving a similar problem. a,b,c are roots of x^3 −7x^2 +4x+1=0 so we have: a+b+c=7 ab+bc+ca=4 (is not needed) abc=−1 similarly from (1/x^3 )+(4/x^2 )−(7/x)+1=0 we have: (1/a)+(1/b)+(1/c)=−4 (1/(ab))+(1/(bc))+(1/(ca))=−7 (is not needed) (1/(abc))=−1 let 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+b+c−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((abc))^(1/3) ((a)^(1/3) +(b)^(1/3) +(c)^(1/3) )((1/( (a)^(1/3) ))+(1/( (b)^(1/3) ))+(1/( (c)^(1/3) ))) ⇒u^3 =7+3−3uv ...(i) 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/( ((ab))^(1/3) ))+(1/( ((bc))^(1/3) ))+(1/( ((ca))^(1/3) ))) v^3 =(1/a)+(1/b)+(1/c)−(3/( ((abc))^(1/3) ))+(3/( ((abc))^(1/3) ))((1/( (a)^(1/3) ))+(1/( (b)^(1/3) ))+(1/( (c)^(1/3) )))((a)^(1/3) +(b)^(1/3) +(c)^(1/3) ) ⇒v^3 =−4+3−3vu ...(ii) (i)×(ii): (uv)^3 =−10−27(uv)+9(uv)^2 (uv)^3 −9(uv)^2 +27(uv)+10=0 let w=uv w^3 −9w^2 +27w+10=0 let w=s+3 (s+3)^3 −9(s+3)^2 +27(s+3)+10=0 s^3 +9s^2 +27s+27−9s^2 −54s−81+27s+81+10=0 s^3 =−37 ⇒s=−((37))^(1/3) ⇒w=uv=3−((37))^(1/3) u^3 =10−3(3−((37))^(1/3) )=1+3((37))^(1/3) ⇒u=(a)^(1/3) +(b)^(1/3) +(c)^(1/3) =((3((37))^(1/3) +1))^(1/3) ✓ v^3 =−1−3(3−((37))^(1/3) )=−10+3((37))^(1/3) ⇒v=(1/( (a)^(1/3) ))+(1/( (b)^(1/3) ))+(1/( (c)^(1/3) ))=((3((37))^(1/3) −10))^(1/3) ✓
$${once}\:{i}\:{developed}\:{following}\:{method} \\ $$$${while}\:{solving}\:{a}\:{similar}\:{problem}. \\ $$$$ \\ $$$${a},{b},{c}\:{are}\:{roots}\:{of}\: \\ $$$$\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{7}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{{x}}+\mathrm{1}=\mathrm{0} \\ $$$${so}\:{we}\:{have}: \\ $$$${a}+{b}+{c}=\mathrm{7} \\ $$$${ab}+{bc}+{ca}=\mathrm{4}\:\:\:\left({is}\:{not}\:{needed}\right) \\ $$$${abc}=−\mathrm{1} \\ $$$${similarly}\:{from}\: \\ $$$$\frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{3}} }+\frac{\mathrm{4}}{\boldsymbol{{x}}^{\mathrm{2}} }−\frac{\mathrm{7}}{\boldsymbol{{x}}}+\mathrm{1}=\mathrm{0} \\ $$$${we}\:{have}: \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=−\mathrm{4} \\ $$$$\frac{\mathrm{1}}{{ab}}+\frac{\mathrm{1}}{{bc}}+\frac{\mathrm{1}}{{ca}}=−\mathrm{7}\:\:\:\left({is}\:{not}\:{needed}\right) \\ $$$$\frac{\mathrm{1}}{{abc}}=−\mathrm{1} \\ $$$${let}\:\boldsymbol{{u}}=\sqrt[{\mathrm{3}}]{\boldsymbol{{a}}}+\sqrt[{\mathrm{3}}]{\boldsymbol{{b}}}+\sqrt[{\mathrm{3}}]{\boldsymbol{{c}}},\: \\ $$$$\:\:\:\:\:\:\boldsymbol{{v}}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{a}}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{b}}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{c}}}} \\ $$$${u}^{\mathrm{3}} ={a}+{b}+{c}−\mathrm{3}\sqrt[{\mathrm{3}}]{{abc}}+\mathrm{3}\left(\sqrt[{\mathrm{3}}]{{a}}+\sqrt[{\mathrm{3}}]{{b}}+\sqrt[{\mathrm{3}}]{{c}}\right)\left(\sqrt[{\mathrm{3}}]{{ab}}+\sqrt[{\mathrm{3}}]{{bc}}+\sqrt[{\mathrm{3}}]{{ca}}\right) \\ $$$${u}^{\mathrm{3}} ={a}+{b}+{c}−\mathrm{3}\sqrt[{\mathrm{3}}]{{abc}}+\mathrm{3}\sqrt[{\mathrm{3}}]{{abc}}\left(\sqrt[{\mathrm{3}}]{{a}}+\sqrt[{\mathrm{3}}]{{b}}+\sqrt[{\mathrm{3}}]{{c}}\right)\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{c}}}\right) \\ $$$$\Rightarrow{u}^{\mathrm{3}} =\mathrm{7}+\mathrm{3}−\mathrm{3}{uv}\:\:\:…\left({i}\right) \\ $$$${v}^{\mathrm{3}} =\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}−\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}}]{{abc}}}+\mathrm{3}\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{c}}}\right)\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{ab}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{bc}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{ca}}}\right) \\ $$$${v}^{\mathrm{3}} =\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}−\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}}]{{abc}}}+\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}}]{{abc}}}\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{c}}}\right)\left(\sqrt[{\mathrm{3}}]{{a}}+\sqrt[{\mathrm{3}}]{{b}}+\sqrt[{\mathrm{3}}]{{c}}\right) \\ $$$$\Rightarrow{v}^{\mathrm{3}} =−\mathrm{4}+\mathrm{3}−\mathrm{3}{vu}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$$\left({uv}\right)^{\mathrm{3}} =−\mathrm{10}−\mathrm{27}\left({uv}\right)+\mathrm{9}\left({uv}\right)^{\mathrm{2}} \\ $$$$\left({uv}\right)^{\mathrm{3}} −\mathrm{9}\left({uv}\right)^{\mathrm{2}} +\mathrm{27}\left({uv}\right)+\mathrm{10}=\mathrm{0} \\ $$$${let}\:{w}={uv} \\ $$$${w}^{\mathrm{3}} −\mathrm{9}{w}^{\mathrm{2}} +\mathrm{27}{w}+\mathrm{10}=\mathrm{0} \\ $$$${let}\:{w}={s}+\mathrm{3} \\ $$$$\left({s}+\mathrm{3}\right)^{\mathrm{3}} −\mathrm{9}\left({s}+\mathrm{3}\right)^{\mathrm{2}} +\mathrm{27}\left({s}+\mathrm{3}\right)+\mathrm{10}=\mathrm{0} \\ $$$${s}^{\mathrm{3}} +\mathrm{9}{s}^{\mathrm{2}} +\mathrm{27}{s}+\mathrm{27}−\mathrm{9}{s}^{\mathrm{2}} −\mathrm{54}{s}−\mathrm{81}+\mathrm{27}{s}+\mathrm{81}+\mathrm{10}=\mathrm{0} \\ $$$${s}^{\mathrm{3}} =−\mathrm{37} \\ $$$$\Rightarrow{s}=−\sqrt[{\mathrm{3}}]{\mathrm{37}} \\ $$$$\Rightarrow{w}={uv}=\mathrm{3}−\sqrt[{\mathrm{3}}]{\mathrm{37}} \\ $$$${u}^{\mathrm{3}} =\mathrm{10}−\mathrm{3}\left(\mathrm{3}−\sqrt[{\mathrm{3}}]{\mathrm{37}}\right)=\mathrm{1}+\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{37}} \\ $$$$\Rightarrow{u}=\sqrt[{\mathrm{3}}]{\boldsymbol{{a}}}+\sqrt[{\mathrm{3}}]{\boldsymbol{{b}}}+\sqrt[{\mathrm{3}}]{\boldsymbol{{c}}}=\sqrt[{\mathrm{3}}]{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{37}}+\mathrm{1}}\:\:\checkmark \\ $$$${v}^{\mathrm{3}} =−\mathrm{1}−\mathrm{3}\left(\mathrm{3}−\sqrt[{\mathrm{3}}]{\mathrm{37}}\right)=−\mathrm{10}+\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{37}} \\ $$$$\Rightarrow{v}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{a}}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{b}}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{c}}}}=\sqrt[{\mathrm{3}}]{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{37}}−\mathrm{10}}\:\:\checkmark \\ $$
Commented by Frix last updated on 24/Jul/25
Great! (“−” missing in the last line)
$$\mathrm{Great}! \\ $$$$\left(“−''\:\mathrm{missing}\:\mathrm{in}\:\mathrm{the}\:\mathrm{last}\:\mathrm{line}\right) \\ $$
Commented by mr W last updated on 25/Jul/25
thanks sir! but i think (1/( (a)^(1/3) ))+(1/( (b)^(1/3) ))+(1/( (c)^(1/3) )) = ((3((37))^(1/3) −10))^(1/3) (1/( (a)^(1/3) ))+(1/( (b)^(1/3) ))+(1/( (c)^(1/3) )) ≠ −((3((37))^(1/3) −10))^(1/3)
$${thanks}\:{sir}! \\ $$$${but}\:{i}\:{think} \\ $$$$\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{a}}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{b}}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{c}}}}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{37}}−\mathrm{10}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{a}}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{b}}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\boldsymbol{{c}}}}\:\neq\:−\sqrt[{\mathrm{3}}]{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{37}}−\mathrm{10}} \\ $$
Commented by Frix last updated on 24/Jul/25
Approximating (just to test this) gives (1/( (a)^(1/3) ))+(1/( (b)^(1/3) ))+(1/( (c)^(1/3) ))≈−.149 and ((3((37))^(1/3) −10))^(1/3) ≈−.149 ⇒ you are right.
$$\mathrm{Approximating}\:\left(\mathrm{just}\:\mathrm{to}\:\mathrm{test}\:\mathrm{this}\right)\:\mathrm{gives} \\ $$$$\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{c}}}\approx−.\mathrm{149} \\ $$$$\mathrm{and} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{37}}−\mathrm{10}}\approx−.\mathrm{149} \\ $$$$\Rightarrow\:\mathrm{you}\:\mathrm{are}\:\mathrm{right}. \\ $$
Commented by mr W last updated on 25/Jul/25
thanks for checking!
$${thanks}\:{for}\:{checking}! \\ $$

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