solve for x∈R x^3 −3((3x−2))^(1/3) +2=0


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Question Number 200739 by mr W last updated on 22/Nov/23

$s o l v e f o r x \in R$ $x^{3} - 3 \sqrt[3]{3 x - 2} + 2 = 0$
Commented by Frix last updated on 23/Nov/23

$x^{3} - p x + q = 0$ $\Leftrightarrow$ $x = \sqrt[3]{p x - q} \land x = \frac{x^{3} + q}{p}$ $x^{3} - p \sqrt[3]{p x - q} + q = 0$ $Nice real solutions with$ $p = 3 a^{2} + b^{2} \land q = - 2 a (a^{2} - b^{2}); a, b \in R$ $\Rightarrow x = 2 a \lor x = - a \pm b$
Commented by Frix last updated on 23/Nov/23

$The behaviour in C is more interesting . . .$ $(using \sqrt[3]{z} = \sqrt[3]{r e^{i θ}} = \sqrt[3]{r} e^{i \frac{θ}{3}} not \sqrt[3]{- r} = - \sqrt[3]{r})$
	Answered by witcher3 last updated on 22/Nov/23
	$((3x−2))^(1/3) =t x=((t^3 +2)/3) ⇔(((t^3 +2)/3))^3 −3t+2=0 ⇔t^9 +6t^6 +12t^3 −81t+62=p(t)=0 p(1)=0,p(−2)=0 by p′(1)=0 p(t)=(t−1)^2 (t+2)(t^6 +3t^4 +4t^3 +9t^2 +6t+31) t^6 +3t^4 +4t^3 +9t^2 +6t+31 =t^6 +t^4 +6t^2 +2t^4 +2t^2 +4t^3 +t^2 +6t+9+22 2t^4 +2t^2 ≥2(√(4t^6 ))=4∣t^3 ∣ ≥t^6 +t^4 +6t^2 +4∣t^3 ∣+4t^3 +(t+3)^2 +22>22 no root in R t=1⇒x=1 t=−2⇒x=−2 x∈{1,−2} inR$
	$\sqrt[3]{3 x - 2} = t$ $x = \frac{t^{3} + 2}{3}$ $\Leftrightarrow {(\frac{t^{3} + 2}{3})}^{3} - 3 t + 2 = 0$ $\Leftrightarrow t^{9} + {6 t}^{6} + {12 t}^{3} - 81 t + 62 = p (t) = 0$ $p (1) = 0, p (- 2) = 0 by$ $p^{'} (1) = 0$ $p (t) = {(t - 1)}^{2} (t + 2) (t^{6} + {3 t}^{4} + {4 t}^{3} + {9 t}^{2} + 6 t + 31)$ $t^{6} + {3 t}^{4} + {4 t}^{3} + {9 t}^{2} + 6 t + 31$ $= t^{6} + t^{4} + {6 t}^{2} + {2 t}^{4} + {2 t}^{2} + {4 t}^{3} + t^{2} + 6 t + 9 + 22$ ${2 t}^{4} + {2 t}^{2} ⩾ 2 \sqrt{{4 t}^{6}} = 4 ∣ t^{3} ∣$ $⩾ t^{6} + t^{4} + {6 t}^{2} + 4 ∣ t^{3} ∣ + {4 t}^{3} + {(t + 3)}^{2} + 22 > 22$ $no root in R$ $t = 1 \Rightarrow x = 1$ $t = - 2 \Rightarrow x = - 2$ $x \in {1, - 2} in R$
	Commented by mr W last updated on 22/Nov/23

	$t h a n k s s i r!$
	Commented by witcher3 last updated on 22/Nov/23

	$withe Pleasur$
	Answered by Frix last updated on 23/Nov/23

	$x, y \in R \Rightarrow using \sqrt[3]{- r} = - \sqrt[3]{r}$ $\frac{x^{3} + 2}{3} = \sqrt[3]{3 x - 2}$ $y_{1} = y_{2}$ $y_{1} = \frac{x^{3} + 2}{3}$ $y_{2} = \sqrt[3]{3 x - 2} \Leftrightarrow x = \frac{y_{2}^{3} + 2}{3}$ $\Rightarrow y = x$ $x = \frac{x^{3} + 2}{3}$ $x^{3} - 3 x + 2 = 0$ $(x + 2) {(x - 1)}^{2} = 0$ $x = - 2 \lor x = 1$
	Commented by mr W last updated on 23/Nov/23

	$t h a n k y o u s i r!$
	Answered by mr W last updated on 23/Nov/23

	$\sqrt[3]{3 x - 2} = \frac{x^{3} + 2}{3}$ $f (x) = y = \sqrt[3]{3 x - 2}$ $\Rightarrow x = \frac{y^{3} + 2}{3}$ $\Rightarrow f^{- 1} (x) = \frac{x^{3} + 2}{3} = \sqrt[3]{3 x - 2} = f (x)$ $\Rightarrow f^{- 1} (x) = f (x) = x$ $\Rightarrow \frac{x^{3} + 2}{3} = x$ $\Rightarrow x^{3} - 3 x + 2 = 0$ $\Rightarrow {(x - 1)}^{2} (x + 2) = 0$ $\Rightarrow x = 1, - 2$
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