Roots of $x^3-x+1$

Question

I am trying to find nice explicit formulas for the roots of the polynomial $x^3-x+1$. Is there some clever way to write down the roots in a reasonably easy way? I found the roots, but my expressions involve tedious terms with multiple roots $(\ldots + (\ldots)^{1/2})^{1/3}+( \ldots -(\ldots)^{1/2})^{1/3}$ and fractions, pretty much what Mathematica drops out as well.

Answer 1

Yes, this is a special polynomial, but you still have to use Cardano's formula to express it. Let $x \to -P$, then,

$$P^3-P-1 = 0$$

Its unique real root $P \approx 1.3247$ is called the plastic constant and plays a role similar to the golden ratio. Various expressions for it are,

$$\begin{aligned} P &= \frac{(9+\sqrt{69})^{1/3}+(9-\sqrt{69})^{1/3}}{2^{1/3}\cdot9^{1/3}}\\[2.5mm] &=\sqrt[3]{1+\sqrt[3]{1+\sqrt[3]{1+\dots}}}\\[2.5mm] &=\tfrac{2}{\sqrt{3}}\cos\big(\tfrac{1}{3}\arccos(\tfrac{3\sqrt{3}}{2}\big) \big) \end{aligned}$$

P.S. Incidentally, the plastic constant $P$ can be found in the vertices of the snub icosidodecadodecahedron,

$\hskip2.8in$

$\hskip1.3in$(Image courtesy of wikipedia and Robert Webb's Stella software)