solve-for-x-R-x-3-3-3x-2-1-3-2-0-

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 \dots

(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

\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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