Prove $\sqrt[3]{ab}+\sqrt[3]{ac}+\sqrt[3]{bc} <0$, where $a,b,c$ are roots of $x^3 +3x^2-24x +1 = 0$

Question

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.

Answer 1

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.

Answer 2

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 $X^3+Y^3+Z^3-3XYZ = -24-3=-27$

Answer 3

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 $(-1,0)$.