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Question Number 154677 by mathdanisur last updated on 20/Sep/21

Solve for real numbers: ((x^9 - 256x^3 - 791)/(84x^3 )) = ((4x + 7))^(1/3)

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\frac{\mathrm{x}^{\mathrm{9}} \:-\:\mathrm{256x}^{\mathrm{3}} \:-\:\mathrm{791}}{\mathrm{84x}^{\mathrm{3}} }\:=\:\sqrt[{\mathrm{3}}]{\mathrm{4x}\:+\:\mathrm{7}} \\ $$

Commented by MJS_new last updated on 20/Sep/21

((x^9 −256x−791)/(84x^3 ))=((4x+7))^(1/3) we can solve it... let x=((4x+7))^(1/3) ⇒ x^3 −4x−7=0 (1) ⇒ ((x^9 −256x−791)/(84x^3 ))=x ⇔ x^9 −84x^4 −256x−791=0 (2) (x^3 −4x−7)(x^6 +4x^4 +7x^3 +16x^2 −28x+113)=0 ⇒ the solutions of (1) also solve (2) x∈R ⇒ x=(((7/2)−((√(3201))/(18))))^(1/3) +(((7/2)+((√(3201))/(18))))^(1/3)

$$\frac{{x}^{\mathrm{9}} −\mathrm{256}{x}−\mathrm{791}}{\mathrm{84}{x}^{\mathrm{3}} }=\sqrt[{\mathrm{3}}]{\mathrm{4}{x}+\mathrm{7}} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{it}... \\ $$$$\mathrm{let}\:{x}=\sqrt[{\mathrm{3}}]{\mathrm{4}{x}+\mathrm{7}}\:\Rightarrow\:{x}^{\mathrm{3}} −\mathrm{4}{x}−\mathrm{7}=\mathrm{0}\:\left(\mathrm{1}\right) \\ $$$$\Rightarrow \\ $$$$\frac{{x}^{\mathrm{9}} −\mathrm{256}{x}−\mathrm{791}}{\mathrm{84}{x}^{\mathrm{3}} }={x} \\ $$$$\Leftrightarrow \\ $$$${x}^{\mathrm{9}} −\mathrm{84}{x}^{\mathrm{4}} −\mathrm{256}{x}−\mathrm{791}=\mathrm{0}\:\left(\mathrm{2}\right) \\ $$$$\left({x}^{\mathrm{3}} −\mathrm{4}{x}−\mathrm{7}\right)\left({x}^{\mathrm{6}} +\mathrm{4}{x}^{\mathrm{4}} +\mathrm{7}{x}^{\mathrm{3}} +\mathrm{16}{x}^{\mathrm{2}} −\mathrm{28}{x}+\mathrm{113}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{the}\:\mathrm{solutions}\:\mathrm{of}\:\left(\mathrm{1}\right)\:\mathrm{also}\:\mathrm{solve}\:\left(\mathrm{2}\right) \\ $$$${x}\in\mathbb{R}\:\Rightarrow\:{x}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{7}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3201}}}{\mathrm{18}}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{7}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3201}}}{\mathrm{18}}} \\ $$

Commented by mathdanisur last updated on 20/Sep/21

There was a typo. Thank you for your attention Ser. The correct is ...256x, not 256x^3 . Sorry for my mistake Ser.

$$\mathrm{There}\:\mathrm{was}\:\mathrm{a}\:\mathrm{typo}.\:\mathrm{Thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{your}\:\mathrm{attention}\:\boldsymbol{\mathrm{S}}\mathrm{er}. \\ $$$$\mathrm{The}\:\mathrm{correct}\:\mathrm{is}\:...\mathrm{256x},\:\mathrm{not}\:\mathrm{256x}^{\mathrm{3}} . \\ $$$$\mathrm{Sorry}\:\mathrm{for}\:\mathrm{my}\:\mathrm{mistake}\:\boldsymbol{\mathrm{S}}\mathrm{er}. \\ $$

Commented by MJS_new last updated on 20/Sep/21

only approximation is possible I get x_1 ≈−2.05382774 x_2 ≈−1.30927779 x_3 ≈2.82876252

$$\mathrm{only}\:\mathrm{approximation}\:\mathrm{is}\:\mathrm{possible} \\ $$$$\mathrm{I}\:\mathrm{get} \\ $$$${x}_{\mathrm{1}} \approx−\mathrm{2}.\mathrm{05382774} \\ $$$${x}_{\mathrm{2}} \approx−\mathrm{1}.\mathrm{30927779} \\ $$$${x}_{\mathrm{3}} \approx\mathrm{2}.\mathrm{82876252} \\ $$

Commented by mathdanisur last updated on 20/Sep/21

creativ solution, thank you Ser

$$\mathrm{creativ}\:\mathrm{solution},\:\mathrm{thank}\:\mathrm{you}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$

Commented by mr W last updated on 20/Sep/21

it′s each time a wonder to me how you get (x^3 −4x−7)(x^6 +4x^4 +7x^3 +16x^2 −28x+113)=0 from x^9 −84x^4 −256x−791=0. great!

$${it}'{s}\:{each}\:{time}\:{a}\:{wonder}\:{to}\:{me}\:{how} \\ $$$${you}\:{get} \\ $$$$\left({x}^{\mathrm{3}} −\mathrm{4}{x}−\mathrm{7}\right)\left({x}^{\mathrm{6}} +\mathrm{4}{x}^{\mathrm{4}} +\mathrm{7}{x}^{\mathrm{3}} +\mathrm{16}{x}^{\mathrm{2}} −\mathrm{28}{x}+\mathrm{113}\right)=\mathrm{0} \\ $$$${from} \\ $$$${x}^{\mathrm{9}} −\mathrm{84}{x}^{\mathrm{4}} −\mathrm{256}{x}−\mathrm{791}=\mathrm{0}. \\ $$$${great}! \\ $$

Commented by mathdanisur last updated on 20/Sep/21

Thank you so much Ser

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$

Commented by MJS_new last updated on 20/Sep/21

this time I first plotted both sides of the given equation and it looked as if the only solution was on the line y=x. then I tried to verify this and it was easy to divide (x^9 −84x^4 −256x−791)/(x^3 −4x−7) I see no chance to get the exact result without this very lucky coincidence...

$$\mathrm{this}\:\mathrm{time}\:\mathrm{I}\:\mathrm{first}\:\mathrm{plotted}\:\mathrm{both}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{given}\:\mathrm{equation}\:\mathrm{and}\:\mathrm{it}\:\mathrm{looked}\:\mathrm{as}\:\mathrm{if}\:\mathrm{the}\:\mathrm{only} \\ $$$$\mathrm{solution}\:\mathrm{was}\:\mathrm{on}\:\mathrm{the}\:\mathrm{line}\:{y}={x}.\:\mathrm{then}\:\mathrm{I}\:\mathrm{tried}\:\mathrm{to} \\ $$$$\mathrm{verify}\:\mathrm{this}\:\mathrm{and}\:\mathrm{it}\:\mathrm{was}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{divide}\: \\ $$$$\left({x}^{\mathrm{9}} −\mathrm{84}{x}^{\mathrm{4}} −\mathrm{256}{x}−\mathrm{791}\right)/\left({x}^{\mathrm{3}} −\mathrm{4}{x}−\mathrm{7}\right) \\ $$$$\mathrm{I}\:\mathrm{see}\:\mathrm{no}\:\mathrm{chance}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{exact}\:\mathrm{result} \\ $$$$\mathrm{without}\:\mathrm{this}\:\mathrm{very}\:\mathrm{lucky}\:\mathrm{coincidence}... \\ $$

Commented by MJS_new last updated on 21/Sep/21

we can build similar questions this way: ((x^9 −a^4 x−(a^3 +b^2 )b)/(3abx^3 ))=((ax+b))^(1/3) it′s more a magic trick than serious math let ((ax+b))^(1/3) =x ⇔ x^3 −ax−b=0 (1) ⇒ ((x^9 −a^4 x−(a^3 +b^2 )b)/(3abx^3 ))=x x^9 −4abx^4 −a^4 x−(a^3 +b^2 )b=0 (2) (x^3 −ax−b)(x^6 +ax^4 +bx^3 +a^2 x^2 −abx+a^3 +b^2 )=0 obviously the real solution(s) of (1) also solve (2).

$$\mathrm{we}\:\mathrm{can}\:\mathrm{build}\:\mathrm{similar}\:\mathrm{questions}\:\mathrm{this}\:\mathrm{way}: \\ $$$$\frac{{x}^{\mathrm{9}} −{a}^{\mathrm{4}} {x}−\left({a}^{\mathrm{3}} +{b}^{\mathrm{2}} \right){b}}{\mathrm{3}{abx}^{\mathrm{3}} }=\sqrt[{\mathrm{3}}]{{ax}+{b}} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{more}\:\mathrm{a}\:\mathrm{magic}\:\mathrm{trick}\:\mathrm{than}\:\mathrm{serious}\:\mathrm{math} \\ $$$$\mathrm{let}\:\sqrt[{\mathrm{3}}]{{ax}+{b}}={x}\:\Leftrightarrow\:{x}^{\mathrm{3}} −{ax}−{b}=\mathrm{0}\:\left(\mathrm{1}\right) \\ $$$$\Rightarrow \\ $$$$\frac{{x}^{\mathrm{9}} −{a}^{\mathrm{4}} {x}−\left({a}^{\mathrm{3}} +{b}^{\mathrm{2}} \right){b}}{\mathrm{3}{abx}^{\mathrm{3}} }={x} \\ $$$${x}^{\mathrm{9}} −\mathrm{4}{abx}^{\mathrm{4}} −{a}^{\mathrm{4}} {x}−\left({a}^{\mathrm{3}} +{b}^{\mathrm{2}} \right){b}=\mathrm{0}\:\left(\mathrm{2}\right) \\ $$$$\left({x}^{\mathrm{3}} −{ax}−{b}\right)\left({x}^{\mathrm{6}} +{ax}^{\mathrm{4}} +{bx}^{\mathrm{3}} +{a}^{\mathrm{2}} {x}^{\mathrm{2}} −{abx}+{a}^{\mathrm{3}} +{b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\mathrm{obviously}\:\mathrm{the}\:\mathrm{real}\:\mathrm{solution}\left(\mathrm{s}\right)\:\mathrm{of}\:\left(\mathrm{1}\right)\:\mathrm{also} \\ $$$$\mathrm{solve}\:\left(\mathrm{2}\right). \\ $$

Commented by MJS_new last updated on 21/Sep/21

question 154303 works exactly the same way

$$\mathrm{question}\:\mathrm{154303}\:\mathrm{works}\:\mathrm{exactly}\:\mathrm{the}\:\mathrm{same}\:\mathrm{way} \\ $$

Commented by mathdanisur last updated on 21/Sep/21

Very nice Ser, thank you

$$\mathrm{Very}\:\mathrm{nice}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thank}\:\mathrm{you} \\ $$

Answered by imjagoll last updated on 21/Sep/21

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