What is the solution of $x^3+x=1$? 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?
 A: $$  \left( \frac{1}{2} \left( 1 + \sqrt{\frac{31}{27}} \right) \right)^{1/3} + \left( \frac{1}{2} \left( 1 - \sqrt{\frac{31}{27}} \right) \right)^{1/3}  $$
A: Alternatively
$$x=\frac2{\sqrt3}\sinh\left( \frac13\sinh^{-1}\frac{3\sqrt3}2\right)$$
which may be less monstrous.
A: 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.
$$x^{2} + 1 = \frac1x$$" />
A: 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.
