<s>[INST] 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.
 [/INST] Note $X^3+Y^3+Z^3 - 3XYZ = (X+Y+Z)(X^2+Y^2+Z^2-XY-YZ-ZX)$.
Using $X^3=ab, Y^3=bc, Z^3 = ca$, as $X, Y, Z$ are distinct (why?), $X^2+Y^2+Z^2> XY+YZ+ZX$ and hence the sign of $X+Y+Z = \sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ca}$ is the same as the sign of $X^3+Y^3+Z^3-3XYZ = -24-3=-27$
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