Determining whether there are solutions to the cubic polynomial equation $x^3 - x = k - k^3$ other than $x = -k$ for a given parameter $k$ Let $k$ be a real parameter, and consider the equation

$$x^3 - x = k - k^3 .$$

Obviously, $x=-k$ is a solution. Is it the only one? How to prove it?

Rearranging gives $$x + k = x^3 + k^3 = (x + k) (x^2 - x k + k^2),$$ and for any solution other than $x = -k$, we can divide both sides by $x + k$, leaving
$$x^2 - x k + k^2 = 1.$$
As a function of $(x, k)$ the left-hand side is a positive-definite quadratic form, so the solution set $(x, k)$ of this latter equation over $\Bbb R$ is a (nondegenerate) ellipse in the $xk$-plane. So, whether there are solutions other than $x = -k$ depends on whether the line corresponding to the value of the parameter $k$ intersects this ellipse or not.
We now use some elementary calculus to find for which values it does: If we regard $k$ as an implicit function of $x$, differentiating gives
$$2 x - \left(x \frac{dk}{dx} + k\right) + 2 k \frac{dk}{dx} = 0 ,$$ and where $k$ takes on an extremal value we have $\frac{dk}{dx} = 0$. Substituting and rearranging gives $x = \frac{k}{2} ,$ and substituting in the ellipse equation gives
$$\left(\frac{k}{2}\right)^2 - \left(\frac{k}{2}\right) k + k^2 = 1.$$ Simplifying gives $$\frac{3 k^2}{4} = 1,$$ and solving gives the extremal values of $k$ on the ellipse, that is, the extremal values of $k$ for which there is a solution other than $x = -k$, namely, $x = \pm \frac{2}{\sqrt{3}}$. (Strictly we must check that such solutions are not all on the line $x = -k$, but this is immediately clear from a graph, and is anyway not hard to see algebraically.) We conclude that there is a solution to the equation other than $x = -k$ if the parameter $k$ satisfies
$$-\frac{2}{\sqrt{3}} \leq k \leq \frac{2}{\sqrt{3}} .$$
