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

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The problem presented involves proving that the sum of the cube roots of the roots of the cubic equation x^3 + 3x^2 - 24x + 1 equals zero. The roots of the equation are denoted as a, b, and c. Participants in the discussion provided various approaches to demonstrate this relationship, ultimately leading to a consensus on the solution. Notable contributors who successfully solved the problem include Olinguito, Opalg, kaliprasad, and lfdahl. The discussion emphasizes the importance of understanding the properties of cubic equations and their roots.
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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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Congratulations to the following members for their correct solution!(Cool)

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

Solution from Opalg:
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, $\sqrt[3]a + \sqrt[3]b + \sqrt[3]c = 0$.
 
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