<s>[INST] 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.
 [/INST] Hint $\;-\;$ using $\,\alpha\beta+\beta\gamma+\gamma\alpha=-3\,$:
$$
\require{cancel}
\begin{align}
m^2 &= \left(|\alpha|+|\beta|+|\gamma|\right)^2
\\ &= \alpha^2+\beta^2+\gamma^2 + 2\left(|\alpha\beta|+|\beta\gamma|+|\gamma\alpha|\right)
\\ &= \cancel{\left(\alpha+\beta+\gamma\right)^2} - 2(\alpha\beta+\beta\gamma+\gamma\alpha)+ 2\left(|\alpha\beta|+|\beta\gamma|+|\gamma\alpha|\right)
\\ &= 6 + 2\left(|\alpha\beta|+|\beta\gamma|+|\gamma\alpha|\right)
\\ &\ge 6 + 2\,|\alpha\beta+\beta\gamma+\gamma\alpha|
\\ &= 12
\end{align}
$$
This gives a lower bound on $m^2\,$. To prove it's an actual minimum, it is enough to find a $k$ such that $|\alpha|+|\beta|+|\gamma| = 2 \sqrt{3}\,$, which turns out not to be too hard.
Note: the above assumes the roots are real (per OP's comments), in order for $\,|\alpha|^2=\alpha^2\,$ to hold.
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