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 assuming that the question intends to ask that each of $\{\alpha,\beta,\gamma\}$ must be real in addition to $k$. If they are allowed to be non-real, the situation is trickier.
The sum $|\alpha|+|\beta|+|\gamma|$ is preserved if $k$ is replaced by $-k$, so we can without loss of generality assume that $k<0$, so that there are two positive roots and one negative root. Let the positive roots be $\alpha>\beta$, and let $\alpha(k)$ and $\beta(k)$ represent the functions mapping $k$ to the largest root and to the second largest root of $x^3/3-x-k$, respectively. We seek to minimize $2\alpha(k)+2\beta(k)$.
See if you can do this via calculus, writing the derivative of $2\alpha(k)+2\beta(k)$ in terms of $\alpha(k)$, $\beta(k)$, and $k$. Then you only need to check the endpoints ($k=0$ and $k=-2/3$), and any places where the derivative may be $0$.
