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According to Wolfram|Alpha, the solution of $x^3+x=1$ is approximate $0.68233$ or exactly this monstrosity:

$x_0=\frac{\sqrt[3]{\frac{1}{2}(9+\sqrt{93})}}{3^{\frac{2}{3}}}-\sqrt[3]{\frac{2}{3(9+\sqrt{93})}}$

$x^3+x=1$ is so simple, that I refuse to believe that this ugly construct is the simplest way. Am I right?

zvavybir
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    Then ask Wolfram alpha for exact forms for the just mildly harder $x^4+x=1$. Not to mention $x^5+x^2+x=1$ – Hagen von Eitzen Nov 21 '20 at 21:18
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    No, @ErnestBredar... you are simply wrong. – David G. Stork Nov 21 '20 at 21:20
  • I see no a priori reason to expect a much nicer answer. Although it is a known fact that WA some times misses simplifications, most of the time they are on the mark. – Arthur Nov 21 '20 at 21:20
  • @HagenvonEitzen Holy Shit… I understand what you mean. – zvavybir Nov 21 '20 at 21:22
  • @DavidG.Stork too bad!! – zvavybir Nov 21 '20 at 21:23
  • @Arthur Mathematics is beautiful and that means that simple things don't get ugly. – zvavybir Nov 21 '20 at 21:24
  • @Ernest Bredar, As David alluded to, there's no reason to think these equations are simple. You just think that because they can written down easily. – Derek Luna Nov 21 '20 at 21:26
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    ``Simple'' is subjective. One can argue that Fermat's last theorem is simple, but it took what, 300 years to prove it. – chhro Nov 21 '20 at 21:26
  • well $x^2+1=0$ is also simple, yet $i$ is a monstrosity (don't be lured by its cute name), just examine the pandora box it opened... – zwim Nov 21 '20 at 21:35
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    @zwim Why do you think $i$ is a monstrosity? It is in no way other than $-1$ or $\frac{1}{2}$. – zvavybir Nov 21 '20 at 21:37
  • Change the exponent $3$, by $7$ for example, and you will still have an expression "so simple" but the result will be much uglier, believe me. – Piquito Nov 21 '20 at 21:51
  • @Ernest Bredar you're in for quite a shock then. –  Nov 21 '20 at 23:28
  • @Ernest Bredar: That is not how logic works. The laws of logic do not care about the nonsensical, human, and emotional concept of beauty, which is also entirely subjective and vague. Simplicity, on the same vein, is also vague and subjective. Unfortunately for you, there is no law of logic that states that simple postulations must have beautiful, simple, or satisfying answers. This is the truth. – Angel Nov 22 '20 at 06:32

4 Answers4

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$$ \left( \frac{1}{2} \left( 1 + \sqrt{\frac{31}{27}} \right) \right)^{1/3} + \left( \frac{1}{2} \left( 1 - \sqrt{\frac{31}{27}} \right) \right)^{1/3} $$

Will Jagy
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Alternatively $$x=\frac2{\sqrt3}\sinh\left( \frac13\sinh^{-1}\frac{3\sqrt3}2\right)$$ which may be less monstrous.

Quanto
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If one re-arranges $x^{3}+x= 1$ as

$x^{2} + 1 = \frac1x$

then the accompanying figure shows an alternative approach to obtaining the (real) solution iteratively, starting with some root approximation (e.g. $x_0=0.5$ in the figure), calculating $x_0^{2} + 1$ to obtain the first iteration $x_1=\frac{1}{x_0^{2}+1}$ and so on. In comparison to the unavoidable ways of expressing the exact solution, the iterative equation for convergence on 0.68233 (5 d.p.) looks rather simple:

$x_{n+1} = \frac{1}{x_n^{2}+1}$

Such is the symmetry of the curves (I've only shown the quadrant with real root), there is no constraint on the choice of initial real value $x_0$ to achieve convergence.

<span class=$$x^{2} + 1 = \frac1x$$" />

Jimbo
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You can rearrange x^3 + x = 1 as x^3 = 1 – x, and let x = u + v. Notice that by the binomial theorem, (u + v)^3 = u^3 + 3vu^2 + 3uv^2 + v^3 = u^3 + v^3 + 3uv(u + v), suggesting that u^3 + v^3 = 1, while -1 = 3uv. -1 = 3uv implies v = -1/(3u), so u^3 – 1/(27u^3) = 1, implying 27u^6 – 1 = 27u^3. You can rearrange this as 27(u^3)^2 – 27u^3 – 1 = 0. This is a quadratic with respect to u^3, so you can use the quadratic formula to conclude that u^3 = [27 + sqrt(93)]/54 or u^3 = [27 – sqrt(93)]/54, although it really does not matter, since v^3 will always be the conjugate of u^3. Therefore, x = u + v = cbrt([27 + sqrt(93)]/54) + cbrt([27 + sqrt(93)]/54). This is ultimately equivalent to what was posted, it just requires some algebraic manipulations to get there. And yes, this is the simplest way you can write the answer. It is what it is. Unfortunately, simple problems do not always have simple solutions. And the laws of logic do not care about our puny concept of simplicity anyway.

Angel
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