Cubic equation problem $\frac{x^3}{3}-x=k$ The cubic function $\frac{{{x^3}}}{3} - x = k$ has three different roots $\alpha,\beta,\gamma$ about the real number k. Let's call the minimum value of $|\alpha|+|\beta|+|\gamma|$ as $m$. FInd the value of $m^2$.
My approach is as follow
$\frac{{{x^3}}}{3} - x = k \Rightarrow f\left( x \right) = \frac{{{x^3}}}{3} - x - k$
$f'\left( x \right) = {x^2} - 1 = 0$
Hence $x =  \pm 1$
$f\left( 1 \right) = \frac{1}{3} - 1 - k =  - \left( {k + \frac{2}{3}} \right)\& f\left( { - 1} \right) =  - \frac{1}{3} + 1 - k =  - \left( {k - \frac{2}{3}} \right)$
For real roots $f\left( 1 \right)f\left( { - 1} \right) < 0$
Therefore $ - \left( {k + \frac{2}{3}} \right) \times \left( { - \left( {k - \frac{2}{3}} \right)} \right) < 0$
$k \in \left( { - \frac{2}{3},\frac{2}{3}} \right)$
We know that $\alpha+\beta+\gamma=0$
But how we will find the minimum values of the sum of the modulus of the roots.

Note: this is similar to another answer, except that by recognizing only one root needs to be tracked instead of the other two the calculation is simplified.
First observe that if we reverse the sign of $k$ then all roots are also reversed in sign with no effect on the set of absolute values, thus no impact on the sum $|\alpha|+|\beta|+|\gamma|$. So we can cover all values of this sum by considering just the case where $k$ is nonpositive.
Then only one root is negative by Descartes' Rule of Signs. Calling that root $\alpha$ we then have
$|\alpha|+|\beta|+|\gamma|=-\alpha+\beta+\gamma$
$\color{blue}{=-2\alpha\because\alpha+\beta+\gamma=0}$
Thus find the nonpositive value of $k$ that minimizes the absolute value of the negative root $\alpha$. Given that $k=0$ gives $\alpha=-\sqrt3$ and $k<0$ must give $\alpha<-\sqrt3$ (why?), you then get your answer using the blue equation above.
