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Question Number 71406 by mind is power last updated on 15/Oct/19
Hello Solve (x^2 )^(1/3) −3((x(x−1)))^(1/3) +2((x−1))^(1/3) =0
$$\mathrm{Hello}\: \\ $$$$\mathrm{Solve}\:\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{2}} }−\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{x}\left(\mathrm{x}−\mathrm{1}\right)}+\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{1}}=\mathrm{0} \\ $$
Commented by Prithwish sen last updated on 15/Oct/19
x^2 +(x−1)−27x(x−1)+18x(x−1)^(2/3) =0 putting x =a^3 and x−1 = b^3 ⇒a^3 = 1+b^3 ∴ a^6 +8b^3 −27a^3 b^3 +18a^3 b^2 = 0 ⇒26b^6 −18b^5 +17b^3 −18b^2 −1=0 and I got a equation of 6 degrees. Sir please comment.
$$\:\mathrm{x}^{\mathrm{2}} +\left(\mathrm{x}−\mathrm{1}\right)−\mathrm{27x}\left(\mathrm{x}−\mathrm{1}\right)+\mathrm{18x}\left(\mathrm{x}−\mathrm{1}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} =\mathrm{0} \\ $$$$\boldsymbol{\mathrm{putting}}\:\boldsymbol{\mathrm{x}}\:=\boldsymbol{\mathrm{a}}^{\mathrm{3}} \:\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{x}}−\mathrm{1}\:=\:\boldsymbol{\mathrm{b}}^{\mathrm{3}} \:\:\Rightarrow\boldsymbol{\mathrm{a}}^{\mathrm{3}} \:=\:\mathrm{1}+\boldsymbol{\mathrm{b}}^{\mathrm{3}} \\ $$$$\therefore\:\mathrm{a}^{\mathrm{6}} +\mathrm{8b}^{\mathrm{3}} −\mathrm{27a}^{\mathrm{3}} \mathrm{b}^{\mathrm{3}} +\mathrm{18a}^{\mathrm{3}} \mathrm{b}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{26}\boldsymbol{\mathrm{b}}^{\mathrm{6}} −\mathrm{18}\boldsymbol{\mathrm{b}}^{\mathrm{5}} +\mathrm{17}\boldsymbol{\mathrm{b}}^{\mathrm{3}} −\mathrm{18}\boldsymbol{\mathrm{b}}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{I}}\:\boldsymbol{\mathrm{got}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{equation}}\:\boldsymbol{\mathrm{of}}\:\mathrm{6}\:\boldsymbol{\mathrm{degrees}}. \\ $$$$\boldsymbol{\mathrm{Sir}}\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{comment}}. \\ $$
Commented by MJS last updated on 15/Oct/19
a^(1/3) −b^(1/3) +c^(1/3) =0 a^(1/3) +c^(1/3) =b^(1/3) a+c+3a^(1/3) c^(1/3) (a^(1/3) +c^(1/3) )=b 3a^(1/3) b^(1/3) c^(1/3) =b−a−c 27abc=(b−a−c)^3 a^3 −b^3 +c^3 +21abc−3(a^2 b+a^2 c+ab^2 +ac^2 +b^2 c−bc^2 )=0 a=x^2 b=27x(x−1) c=8(x−1) x^6 −((19203)/(4394))x^5 +((61719)/(8788))x^4 −((92387)/(17576))x^3 +((4299)/(2197))x^2 −((840)/(2197))x+((64)/(2197))=0 I found no exact solution x_1 ≈.172484 x_2 ≈1.72890 x_(3, 4) ≈.230494±.209231i x_(5, 6) ≈1.00395±.0115209i if we generally use (z)^(1/3) =abs (z)^(1/3) ×e^(i((arg (z))/3)) only x_2 is a solution if we use the common exception ((−r))^(1/3) =−(r)^(1/3) also x_1 is a solution the complex solutions are false
$${a}^{\mathrm{1}/\mathrm{3}} −{b}^{\mathrm{1}/\mathrm{3}} +{c}^{\mathrm{1}/\mathrm{3}} =\mathrm{0} \\ $$$${a}^{\mathrm{1}/\mathrm{3}} +{c}^{\mathrm{1}/\mathrm{3}} ={b}^{\mathrm{1}/\mathrm{3}} \\ $$$${a}+{c}+\mathrm{3}{a}^{\mathrm{1}/\mathrm{3}} {c}^{\mathrm{1}/\mathrm{3}} \left({a}^{\mathrm{1}/\mathrm{3}} +{c}^{\mathrm{1}/\mathrm{3}} \right)={b} \\ $$$$\mathrm{3}{a}^{\mathrm{1}/\mathrm{3}} {b}^{\mathrm{1}/\mathrm{3}} {c}^{\mathrm{1}/\mathrm{3}} ={b}−{a}−{c} \\ $$$$\mathrm{27}{abc}=\left({b}−{a}−{c}\right)^{\mathrm{3}} \\ $$$${a}^{\mathrm{3}} −{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +\mathrm{21}{abc}−\mathrm{3}\left({a}^{\mathrm{2}} {b}+{a}^{\mathrm{2}} {c}+{ab}^{\mathrm{2}} +{ac}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}−{bc}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${a}={x}^{\mathrm{2}} \\ $$$${b}=\mathrm{27}{x}\left({x}−\mathrm{1}\right) \\ $$$${c}=\mathrm{8}\left({x}−\mathrm{1}\right) \\ $$$${x}^{\mathrm{6}} −\frac{\mathrm{19203}}{\mathrm{4394}}{x}^{\mathrm{5}} +\frac{\mathrm{61719}}{\mathrm{8788}}{x}^{\mathrm{4}} −\frac{\mathrm{92387}}{\mathrm{17576}}{x}^{\mathrm{3}} +\frac{\mathrm{4299}}{\mathrm{2197}}{x}^{\mathrm{2}} −\frac{\mathrm{840}}{\mathrm{2197}}{x}+\frac{\mathrm{64}}{\mathrm{2197}}=\mathrm{0} \\ $$$$\mathrm{I}\:\mathrm{found}\:\mathrm{no}\:\mathrm{exact}\:\mathrm{solution} \\ $$$${x}_{\mathrm{1}} \approx.\mathrm{172484} \\ $$$${x}_{\mathrm{2}} \approx\mathrm{1}.\mathrm{72890} \\ $$$${x}_{\mathrm{3},\:\mathrm{4}} \approx.\mathrm{230494}\pm.\mathrm{209231i} \\ $$$${x}_{\mathrm{5},\:\mathrm{6}} \approx\mathrm{1}.\mathrm{00395}\pm.\mathrm{0115209i} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{generally}\:\mathrm{use}\:\sqrt[{\mathrm{3}}]{{z}}=\mathrm{abs}\:\left({z}\right)^{\mathrm{1}/\mathrm{3}} ×\mathrm{e}^{\mathrm{i}\frac{\mathrm{arg}\:\left({z}\right)}{\mathrm{3}}} \\ $$$$\mathrm{only}\:{x}_{\mathrm{2}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{solution} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{use}\:\mathrm{the}\:\mathrm{common}\:\mathrm{exception}\:\sqrt[{\mathrm{3}}]{−{r}}=−\sqrt[{\mathrm{3}}]{{r}} \\ $$$$\mathrm{also}\:{x}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{solution} \\ $$$$\mathrm{the}\:\mathrm{complex}\:\mathrm{solutions}\:\mathrm{are}\:\mathrm{false} \\ $$
Commented by Prithwish sen last updated on 15/Oct/19
you just turn the equation in cubic form. Great sir.
$$\mathrm{you}\:\mathrm{just}\:\mathrm{turn}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{in}\:\mathrm{cubic}\:\mathrm{form}. \\ $$$$\mathrm{Great}\:\mathrm{sir}. \\ $$
Commented by mind is power last updated on 15/Oct/19
thank you !, i have no better solution my bee we can′t find exacte root ′ or its hard we need find function g (t) such that if we put x=g(t) equation withe t is easier i try but no result
$$\mathrm{thank}\:\mathrm{you}\:!,\:\:\mathrm{i}\:\mathrm{have}\:\mathrm{no}\:\mathrm{better}\:\mathrm{solution}\:\mathrm{my}\:\mathrm{bee}\:\mathrm{we}\:\mathrm{can}'\mathrm{t}\:\mathrm{find}\:\mathrm{exacte}\:\mathrm{root}\:' \\ $$$$\mathrm{or}\:\mathrm{its}\:\mathrm{hard}\:\mathrm{we}\:\mathrm{need}\:\mathrm{find}\:\mathrm{function}\:\mathrm{g}\:\left(\mathrm{t}\right)\:\mathrm{such}\:\mathrm{that}\:\mathrm{if}\:\mathrm{we}\:\mathrm{put}\:\mathrm{x}=\mathrm{g}\left(\mathrm{t}\right) \\ $$$$\mathrm{equation}\:\mathrm{withe}\:\mathrm{t}\:\mathrm{is}\:\:\mathrm{easier}\:\mathrm{i}\:\mathrm{try}\:\mathrm{but}\:\mathrm{no}\:\mathrm{result}\: \\ $$
Commented by MJS last updated on 15/Oct/19
your equation is equivalent to mine
$$\mathrm{your}\:\mathrm{equation}\:\mathrm{is}\:\mathrm{equivalent}\:\mathrm{to}\:\mathrm{mine} \\ $$
Commented by mind is power last updated on 15/Oct/19
i have did like you sir !
$$\mathrm{i}\:\mathrm{have}\:\mathrm{did}\:\mathrm{like}\:\mathrm{you}\:\mathrm{sir}\:! \\ $$

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