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if-a-b-and-c-root-of-the-x-3-16x-2-57x-1-0-thi-find-thd-volue-of-a-1-5-b-1-5-c-1-5-

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Question Number 190520 by mathlove last updated on 04/Apr/23
if a,b and c root of the x^3 −16x^2 −57x+1=0 thi find thd volue of a^(1/5) +b^(1/5) +c^(1/5) =?
$${if}\:{a},{b}\:{and}\:{c}\:{root}\:{of}\:{the} \\ $$$${x}^{\mathrm{3}} −\mathrm{16}{x}^{\mathrm{2}} −\mathrm{57}{x}+\mathrm{1}=\mathrm{0} \\ $$$${thi}\:{find}\:{thd}\:{volue}\:{of} \\ $$$${a}^{\frac{\mathrm{1}}{\mathrm{5}}} +{b}^{\frac{\mathrm{1}}{\mathrm{5}}} +{c}^{\frac{\mathrm{1}}{\mathrm{5}}} =? \\ $$
Answered by Frix last updated on 05/Apr/23
If the solution is ∈Z we need to find a polynomial factor of degree 3 of t^(15) −16t^(10) −57t^5 +1 I used software to find the factor t^3 −1t^2 −2t+1 The other factor of degree 12 is prime ⇒ answer is 1 [I don′t know if there′s an analytical method to solve this]
$$\mathrm{If}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{is}\:\in\mathbb{Z}\:\mathrm{we}\:\mathrm{need}\:\mathrm{to}\:\mathrm{find}\:\mathrm{a} \\ $$$$\mathrm{polynomial}\:\mathrm{factor}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{3}\:\mathrm{of} \\ $$$${t}^{\mathrm{15}} −\mathrm{16}{t}^{\mathrm{10}} −\mathrm{57}{t}^{\mathrm{5}} +\mathrm{1} \\ $$$$\mathrm{I}\:\mathrm{used}\:\mathrm{software}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{factor} \\ $$$${t}^{\mathrm{3}} −\mathrm{1}{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1} \\ $$$$\mathrm{The}\:\mathrm{other}\:\mathrm{factor}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{12}\:\mathrm{is}\:\mathrm{prime} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{1} \\ $$$$\left[\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{if}\:\mathrm{there}'\mathrm{s}\:\mathrm{an}\:\mathrm{analytical}\right. \\ $$$$\left.\mathrm{method}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{this}\right] \\ $$
Answered by behi834171 last updated on 05/Apr/23
x^3 =16x^2 +57x−1 x^4 =16x^3 +57x^2 −x=16(16x^2 +57x−1)+57x^2 −x= =313x^2 +911x−16 x^5 =313x^3 +911x^2 −16x=313(16x^2 +57x−1)+911x^2 −16x= =5919x^2 +17852x−313 ⇒x^5 −5919x^2 −17852x+313=0 y=(1/x)⇒(1/y^5 )−((5919)/y^2 )−((17852)/y)+313=0 ⇒313y^5 −17852y^4 −5919y^3 +1=0 ⇒Σa^(1/5) =−((−17852)/(313))=((17852)/(313)) .■
$${x}^{\mathrm{3}} =\mathrm{16}{x}^{\mathrm{2}} +\mathrm{57}{x}−\mathrm{1} \\ $$$${x}^{\mathrm{4}} =\mathrm{16}{x}^{\mathrm{3}} +\mathrm{57}{x}^{\mathrm{2}} −{x}=\mathrm{16}\left(\mathrm{16}{x}^{\mathrm{2}} +\mathrm{57}{x}−\mathrm{1}\right)+\mathrm{57}{x}^{\mathrm{2}} −{x}= \\ $$$$=\mathrm{313}{x}^{\mathrm{2}} +\mathrm{911}{x}−\mathrm{16} \\ $$$${x}^{\mathrm{5}} =\mathrm{313}{x}^{\mathrm{3}} +\mathrm{911}{x}^{\mathrm{2}} −\mathrm{16}{x}=\mathrm{313}\left(\mathrm{16}{x}^{\mathrm{2}} +\mathrm{57}{x}−\mathrm{1}\right)+\mathrm{911}{x}^{\mathrm{2}} −\mathrm{16}{x}= \\ $$$$=\mathrm{5919}{x}^{\mathrm{2}} +\mathrm{17852}{x}−\mathrm{313} \\ $$$$\Rightarrow\boldsymbol{{x}}^{\mathrm{5}} −\mathrm{5919}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{17852}\boldsymbol{{x}}+\mathrm{313}=\mathrm{0} \\ $$$${y}=\frac{\mathrm{1}}{{x}}\Rightarrow\frac{\mathrm{1}}{{y}^{\mathrm{5}} }−\frac{\mathrm{5919}}{{y}^{\mathrm{2}} }−\frac{\mathrm{17852}}{{y}}+\mathrm{313}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{313}\boldsymbol{{y}}^{\mathrm{5}} −\mathrm{17852}\boldsymbol{{y}}^{\mathrm{4}} −\mathrm{5919}\boldsymbol{{y}}^{\mathrm{3}} +\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\Sigma\boldsymbol{{a}}^{\frac{\mathrm{1}}{\mathrm{5}}} =−\frac{−\mathrm{17852}}{\mathrm{313}}=\frac{\mathrm{17852}}{\mathrm{313}}\:\:\:\:\:\:.\blacksquare \\ $$
Commented by mr W last updated on 07/Apr/23
thanks sir! your method seems to be a good idea, but i think it is not correct. when you transform the equation to 313y^5 −17852y^4 −5919y^3 +1=0, it has 5 roots. so you get with Σy=Σ(1/x)=((17852)/(313)) the sum of all its 5 roots, i.e. (1/a)+(1/b)+(1/c)+(1/d)+(1/e)=((17852)/(313)). but this is not that what the original question has requested: a^(1/5) +b^(1/5) +c^(1/5) =?
$${thanks}\:{sir}! \\ $$$${your}\:{method}\:{seems}\:{to}\:{be}\:{a}\:{good} \\ $$$${idea},\:{but}\:{i}\:{think}\:{it}\:{is}\:{not}\:{correct}. \\ $$$${when}\:{you}\:{transform}\:{the}\:{equation} \\ $$$${to} \\ $$$$\mathrm{313}\boldsymbol{{y}}^{\mathrm{5}} −\mathrm{17852}\boldsymbol{{y}}^{\mathrm{4}} −\mathrm{5919}\boldsymbol{{y}}^{\mathrm{3}} +\mathrm{1}=\mathrm{0}, \\ $$$${it}\:{has}\:\mathrm{5}\:{roots}.\:{so}\:{you}\:{get}\:{with} \\ $$$$\Sigma{y}=\Sigma\frac{\mathrm{1}}{{x}}=\frac{\mathrm{17852}}{\mathrm{313}}\:{the}\:{sum}\:{of}\:{all}\:{its}\:\mathrm{5} \\ $$$${roots}, \\ $$$${i}.{e}.\:\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\frac{\mathrm{1}}{{d}}+\frac{\mathrm{1}}{{e}}=\frac{\mathrm{17852}}{\mathrm{313}}. \\ $$$${but}\:{this}\:{is}\:{not}\:{that}\:{what}\:{the}\:{original} \\ $$$${question}\:{has}\:{requested}: \\ $$$${a}^{\frac{\mathrm{1}}{\mathrm{5}}} +{b}^{\frac{\mathrm{1}}{\mathrm{5}}} +{c}^{\frac{\mathrm{1}}{\mathrm{5}}} =? \\ $$
Commented by behi834171 last updated on 08/Apr/23
hello my dear master. you are right, and ,i have a terrible typo i dont delete my way,but will correct it and i will post that. thanks for pointing this.
$${hello}\:{my}\:{dear}\:{master}. \\ $$$${you}\:{are}\:{right},\:{and}\:,{i}\:{have}\:{a}\:{terrible}\:{typo} \\ $$$${i}\:{dont}\:{delete}\:\:{my}\:{way},{but}\:{will}\:{correct}\:{it} \\ $$$${and}\:{i}\:{will}\:{post}\:{that}. \\ $$$${thanks}\:{for}\:{pointing}\:{this}. \\ $$

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