I am interested in finding a solution for the equation:
$$ x^3 - x + 1 = 0 $$
I've noticed that there are numerous polynomial equations where one of the coefficients is zero. Could you provide guidance on solving such equations? Additionally, if you could recommend any resources for further exploration, I would greatly appreciate it.
On
May be, you could prefer $$x=-\frac{2}{\sqrt{3}} \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{3\sqrt{3}}{2}\right)\right)$$ obtained following the steps described here
Write $x$ as $u + v$. Then you could write the equation as $(u+v)^3 - (u+v) + 1 = 0$. Expanding it out, you get:
$$u^3 + 3 u^2 v + 3 u v^2 + v^3 - (u + v) + 1 = (u^3 + v^3 + 1) + 3uv(u+v)-(u+v)=0 = (u^3+v^3+1) + (3uv-1)(u+v)=0.$$
Forcing $u^3+v^3+1$ and $3uv-1$ to be $0$, you get that the following must be true:
$$\begin{cases}u^3+v^3=-1 \\ u^3 v^3 = \frac{1}{27} \end{cases}$$
At this point you know the sum and the product of $u^3$ and $v^3$, so you can solve for them through a quadratic equation (or you could just solve the system of equation I wrote earlier):
$$x^2 - (u^3 + v^3)x + u^3 v^3 = 0 \iff x^2 + x + \frac{1}{27},$$
which is solved for $x = -\frac{1}2 \pm \frac{\sqrt{\frac{23}{3}}}6$. Then you get that $u^3$ is equal to one of those roots, while $v^3$ is the other. Then, since $x = u + v$, you get the solution you wanted:
$$x = \sqrt[3]{-\frac{1}2 + \frac{\sqrt{\frac{23}{3}}}6} + \sqrt[3]{-\frac{1}2 - \frac{\sqrt{\frac{23}{3}}}6} \approx -1.3247.$$
Now you can get the other two roots by using Ruffini's rule and solving another quadratic equation. You can generalise this process to solve all the cubics of the form $x^3 + bx + c = 0$, as AlvinL pointed out in a comment -- this is known as Cardano's rule.