Prove the Sum of Cube Roots of Roots of a Cubic Equation is Zero

[anemone]

Here is this week's POTW:

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If $a,b,c$ are roots of the equation $x^3+3x^2-24x+1=0$, prove that $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=0$.

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[anemone]

Congratulations to the following members for their correct solution!(Cool)

1. Olinguito
2. Opalg
3. kaliprasad
4. lfdahl

Solution from Opalg:

Spoiler

Let $x^3 + px^2 + qx + r = 0$ be the equation with roots $\sqrt[3]a$, $\sqrt[3]b$, $\sqrt[3]c$. Then the equation $x + px^{2/3} + qx^{1/3} + r=0$ has roots $a$, $b$, $c$. Write that equation as $x+r = -x^{1/3}( px^{1/3} + q)$, and cube both sides: $$x^3 + 3rx^2 + 3r^2x + r^3 = -x(p^3x + 3pqx^{1/3}(px^{1/3} + q) + q^3) = -x(p^3x - 3pq(x+r) + q^3),$$ $$x^3 + (3r + p^3 - 3pq)x^2 + (3r^2 - 3pqr + q^3)x + r^3 = 0.$$ Compare that with the given equation $x^3 + 3x^2 - 24x + 1 = 0$ to see that $$3r + p^3 - 3pq = 3,\qquad 3r^2 - 3pqr + q^3 = -24, \qquad r^3 = 1.$$ Thus $r=1$, and the other two equations become $$p(p^2-3q) = 0, \qquad q(q^2-3p) = -27.$$ In the first of those equations, if $p^2-3q=0$ then $q = \frac13p^2$ and the second equation becomes $\frac{p^2}3\bigl(\frac{p^4}9 - 3p\bigr) = -27$, so that $p^3(p^3 - 27) = -27^2$. But then $\bigl(p^3 - \frac{27}2\bigr)^2 = -\frac34 \cdot27^2$, which is impossible.

So $p^2-3q\ne0$ and therefore $p=0$ (and hence $q=-3$). Thus the equation with roots $\sqrt[3]a$, $\sqrt[3]b$, $\sqrt[3]c$ is $x^3 -3x + 1 = 0$. The coefficient of $x^2$ is zero and so the sum of the roots is zero. In other words, $\sqrt[3]a + \sqrt[3]b + \sqrt[3]c = 0$.