Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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 !

$$\mathrm{i}\:\mathrm{have}\:\mathrm{did}\:\mathrm{like}\:\mathrm{you}\:\mathrm{sir}\:! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com