Prove $\sqrt[3]{ab}+\sqrt[3]{ac}+\sqrt[3]{bc} <0$, where $a,b,c$ are roots of $x^3 +3x^2-24x +1 = 0$ I am working on this problem:
$a, b$ and $c$ are the roots of $x^3 +3x^2-24x +1 = 0$. Prove that
$$S = \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0$$
After I used Vietta's Formula,expanded $S^3$ and used many algebraic tricks I have
$S[S^2 -3(a'b' +a'c +b'c')] = 0$, wich leads to $S = 0 $ or $ S^2 -3(a'b' +a'c +b'c') = 0$, where $a' =\sqrt[3]{a}$,  $b' =\sqrt[3]{b}$ and  $c' =\sqrt[3]{c}$
So the final step is to prove the impossibility of the second case by proving that
$$a'b' +a'c +b'c' < 0$$
(which I verified numerically). It's here that I have trouble.
I also noticed that I didn't use the fact that $ab+ac+bc = -24$ (given by Vietta's Formula), so I guess it should be used here.

Rewrite the equation  $x^3 +3x^2-24x +1 = 0$ as
$$(x+1)^3-27x=0$$
Use $a^3-b^3=(a-b)(a^2+ab+b^2)$ and $ a^2+ab+b^2>0$ to obtain
$$x-3\sqrt[3]x+1=0
$$
which is a cubic equation in $\sqrt[3]x$. Thus, given its roots $\sqrt[3]{a}, \sqrt[3]{b} $ and $\sqrt[3]{c} $, the Vieta’s formulas produces
$$\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0$$
and
$$\sqrt[3]{ab}+\sqrt[3]{ac}+\sqrt[3]{bc} =-3<0$$
Note that the latter is actually not needed to prove the former.
