<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] Rewrite the inequality as
$$\sqrt[3]{abc\over a}+\sqrt[3]{abc\over b}+\sqrt[3]{abc\over c}\lt0$$
but now note that $abc=-1$, so this simplifies to
$$\sqrt[3]{1\over a}+\sqrt[3]{1\over b}+\sqrt[3]{1\over c}\gt0$$
Next, note that $1/a$, $1/b$, and $1/c$ are roots of $x^3-24x^2+3x+1=0$. Calling this polynomial $P$, we see that $P(-1)\lt0$, $P(0)\gt0$, and $P(1)\lt0$, so its three roots lie in $(-1,0)$, $(0,1)$, and $(1,\infty)$. Their cube roots lie in the same intervals, and so the sum of the cube roots is clearly positive, since the one in $(1,\infty)$ more than offsets the one in $(-1,0)$.
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