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Question Number 200739 by mr W last updated on 22/Nov/23
solve for x∈R x^3 −3((3x−2))^(1/3) +2=0
$${solve}\:{for}\:{x}\in{R} \\ $$$${x}^{\mathrm{3}} −\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{3}{x}−\mathrm{2}}+\mathrm{2}=\mathrm{0} \\ $$
Commented by Frix last updated on 23/Nov/23
x^3 −px+q=0 ⇔ x=((px−q))^(1/3) ∧x=((x^3 +q)/p) x^3 −p((px−q))^(1/3) +q=0 Nice real solutions with p=3a^2 +b^2 ∧q=−2a(a^2 −b^2 ); a, b ∈R ⇒ x=2a∨x=−a±b
$${x}^{\mathrm{3}} −{px}+{q}=\mathrm{0} \\ $$$$\Leftrightarrow \\ $$$${x}=\sqrt[{\mathrm{3}}]{{px}−{q}}\wedge{x}=\frac{{x}^{\mathrm{3}} +{q}}{{p}} \\ $$$${x}^{\mathrm{3}} −{p}\sqrt[{\mathrm{3}}]{{px}−{q}}+{q}=\mathrm{0} \\ $$$$\mathrm{Nice}\:\mathrm{real}\:\mathrm{solutions}\:\mathrm{with} \\ $$$${p}=\mathrm{3}{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \wedge{q}=−\mathrm{2}{a}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right);\:{a},\:{b}\:\in\mathbb{R} \\ $$$$\Rightarrow\:{x}=\mathrm{2}{a}\vee{x}=−{a}\pm{b} \\ $$
Commented by Frix last updated on 23/Nov/23
The behaviour in C is more interesting... (using (z)^(1/3) =((re^(iθ) ))^(1/3) =(r)^(1/3) e^(i(θ/3)) not ((−r))^(1/3) =−(r)^(1/3) )
$$\mathrm{The}\:\mathrm{behaviour}\:\mathrm{in}\:\mathbb{C}\:\mathrm{is}\:\mathrm{more}\:\mathrm{interesting}… \\ $$$$\left(\mathrm{using}\:\sqrt[{\mathrm{3}}]{{z}}=\sqrt[{\mathrm{3}}]{{r}\mathrm{e}^{\mathrm{i}\theta} }=\sqrt[{\mathrm{3}}]{{r}}\mathrm{e}^{\mathrm{i}\frac{\theta}{\mathrm{3}}} \:\mathrm{not}\:\sqrt[{\mathrm{3}}]{−{r}}=−\sqrt[{\mathrm{3}}]{{r}}\right) \\ $$
Answered by witcher3 last updated on 22/Nov/23
((3x−2))^(1/3) =t x=((t^3 +2)/3) ⇔(((t^3 +2)/3))^3 −3t+2=0 ⇔t^9 +6t^6 +12t^3 −81t+62=p(t)=0 p(1)=0,p(−2)=0 by p′(1)=0 p(t)=(t−1)^2 (t+2)(t^6 +3t^4 +4t^3 +9t^2 +6t+31) t^6 +3t^4 +4t^3 +9t^2 +6t+31 =t^6 +t^4 +6t^2 +2t^4 +2t^2 +4t^3 +t^2 +6t+9+22 2t^4 +2t^2 ≥2(√(4t^6 ))=4∣t^3 ∣ ≥t^6 +t^4 +6t^2 +4∣t^3 ∣+4t^3 +(t+3)^2 +22>22 no root in R t=1⇒x=1 t=−2⇒x=−2 x∈{1,−2} inR
$$\sqrt[{\mathrm{3}}]{\mathrm{3x}−\mathrm{2}}=\mathrm{t} \\ $$$$\mathrm{x}=\frac{\mathrm{t}^{\mathrm{3}} +\mathrm{2}}{\mathrm{3}} \\ $$$$\Leftrightarrow\left(\frac{\mathrm{t}^{\mathrm{3}} +\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} −\mathrm{3t}+\mathrm{2}=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{t}^{\mathrm{9}} +\mathrm{6t}^{\mathrm{6}} +\mathrm{12t}^{\mathrm{3}} −\mathrm{81t}+\mathrm{62}=\mathrm{p}\left(\mathrm{t}\right)=\mathrm{0} \\ $$$$\mathrm{p}\left(\mathrm{1}\right)=\mathrm{0},\mathrm{p}\left(−\mathrm{2}\right)=\mathrm{0}\:\mathrm{by}\: \\ $$$$\mathrm{p}'\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{p}\left(\mathrm{t}\right)=\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{t}+\mathrm{2}\right)\left(\mathrm{t}^{\mathrm{6}} +\mathrm{3t}^{\mathrm{4}} +\mathrm{4t}^{\mathrm{3}} +\mathrm{9t}^{\mathrm{2}} +\mathrm{6t}+\mathrm{31}\right) \\ $$$$\mathrm{t}^{\mathrm{6}} +\mathrm{3t}^{\mathrm{4}} +\mathrm{4t}^{\mathrm{3}} +\mathrm{9t}^{\mathrm{2}} +\mathrm{6t}+\mathrm{31} \\ $$$$=\mathrm{t}^{\mathrm{6}} +\mathrm{t}^{\mathrm{4}} +\mathrm{6t}^{\mathrm{2}} +\mathrm{2t}^{\mathrm{4}} +\mathrm{2t}^{\mathrm{2}} +\mathrm{4t}^{\mathrm{3}} +\mathrm{t}^{\mathrm{2}} +\mathrm{6t}+\mathrm{9}+\mathrm{22} \\ $$$$\mathrm{2t}^{\mathrm{4}} +\mathrm{2t}^{\mathrm{2}} \geqslant\mathrm{2}\sqrt{\mathrm{4t}^{\mathrm{6}} }=\mathrm{4}\mid\mathrm{t}^{\mathrm{3}} \mid \\ $$$$\geqslant\mathrm{t}^{\mathrm{6}} +\mathrm{t}^{\mathrm{4}} +\mathrm{6t}^{\mathrm{2}} +\mathrm{4}\mid\mathrm{t}^{\mathrm{3}} \mid+\mathrm{4t}^{\mathrm{3}} +\left(\mathrm{t}+\mathrm{3}\right)^{\mathrm{2}} +\mathrm{22}>\mathrm{22} \\ $$$$\mathrm{no}\:\mathrm{root}\:\mathrm{in}\:\mathbb{R} \\ $$$$\mathrm{t}=\mathrm{1}\Rightarrow\mathrm{x}=\mathrm{1} \\ $$$$\mathrm{t}=−\mathrm{2}\Rightarrow\mathrm{x}=−\mathrm{2} \\ $$$$\mathrm{x}\in\left\{\mathrm{1},−\mathrm{2}\right\}\:\mathrm{in}\mathbb{R} \\ $$
Commented by mr W last updated on 22/Nov/23
thanks sir!
$${thanks}\:{sir}! \\ $$
Commented by witcher3 last updated on 22/Nov/23
withe Pleasur
$$\mathrm{withe}\:\mathrm{Pleasur} \\ $$
Answered by Frix last updated on 23/Nov/23
x, y ∈R ⇒ using ((−r))^(1/3) =−(r)^(1/3) ((x^3 +2)/3)=((3x−2))^(1/3) y_1 =y_2 y_1 =((x^3 +2)/3) y_2 =((3x−2))^(1/3) ⇔ x=((y_2 ^3 +2)/3) ⇒ y=x x=((x^3 +2)/3) x^3 −3x+2=0 (x+2)(x−1)^2 =0 x=−2∨x=1
$${x},\:{y}\:\in\mathbb{R}\:\Rightarrow\:\mathrm{using}\:\sqrt[{\mathrm{3}}]{−{r}}=−\sqrt[{\mathrm{3}}]{{r}} \\ $$$$\frac{{x}^{\mathrm{3}} +\mathrm{2}}{\mathrm{3}}=\sqrt[{\mathrm{3}}]{\mathrm{3}{x}−\mathrm{2}} \\ $$$${y}_{\mathrm{1}} ={y}_{\mathrm{2}} \\ $$$${y}_{\mathrm{1}} =\frac{{x}^{\mathrm{3}} +\mathrm{2}}{\mathrm{3}} \\ $$$${y}_{\mathrm{2}} =\sqrt[{\mathrm{3}}]{\mathrm{3}{x}−\mathrm{2}}\:\Leftrightarrow\:{x}=\frac{{y}_{\mathrm{2}} ^{\mathrm{3}} +\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow\:{y}={x} \\ $$$${x}=\frac{{x}^{\mathrm{3}} +\mathrm{2}}{\mathrm{3}} \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{2}=\mathrm{0} \\ $$$$\left({x}+\mathrm{2}\right)\left({x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${x}=−\mathrm{2}\vee{x}=\mathrm{1} \\ $$
Commented by mr W last updated on 23/Nov/23
thank you sir!
$${thank}\:{you}\:{sir}! \\ $$
Answered by mr W last updated on 23/Nov/23
((3x−2))^(1/3) =((x^3 +2)/3) f(x)=y=((3x−2))^(1/3) ⇒x=((y^3 +2)/3) ⇒f^(−1) (x)=((x^3 +2)/3)=((3x−2))^(1/3) =f(x) ⇒f^(−1) (x)=f(x)=x ⇒((x^3 +2)/3)=x ⇒x^3 −3x+2=0 ⇒(x−1)^2 (x+2)=0 ⇒x=1, −2
$$\sqrt[{\mathrm{3}}]{\mathrm{3}{x}−\mathrm{2}}=\frac{{x}^{\mathrm{3}} +\mathrm{2}}{\mathrm{3}} \\ $$$${f}\left({x}\right)={y}=\sqrt[{\mathrm{3}}]{\mathrm{3}{x}−\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{{y}^{\mathrm{3}} +\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\frac{{x}^{\mathrm{3}} +\mathrm{2}}{\mathrm{3}}=\sqrt[{\mathrm{3}}]{\mathrm{3}{x}−\mathrm{2}}={f}\left({x}\right) \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)={f}\left({x}\right)={x} \\ $$$$\Rightarrow\frac{{x}^{\mathrm{3}} +\mathrm{2}}{\mathrm{3}}={x} \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{1},\:−\mathrm{2} \\ $$

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