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How to prove that

$$ \begin{cases} x_1 + x_2 + x_3 = 0 \\ x_1x_2 + x_2x_3 + x_3x_1 = p \\ x_1x_2x_3 = -q \\ x_1 = 1/x_2 + 1/x_3 \end{cases} $$

implies

$$ q^3 + pq + q = 0 $$

?

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1  
Have you tried subbing in your values for p and q and then using the given identities? –  Crockett May 28 at 18:45
1  
First three are Vieta's formula. Not sure about last one. –  Kaster May 28 at 19:02

2 Answers 2

up vote 1 down vote accepted

Denote the equations by $(1),\ldots ,(4)$. Then $(4)$ says $x_1x_2x_3=x_2+x_3$ and $(3)$ says $x_1x_2x_3=-q$. This gives $x_3=-x_2-q$. Substitute this into $(1)$. This gives $x_1=q$. Then $q\cdot (2)-(3)$ gives $-q(p+q^2+1)=0$, or $$q^3+pq+q=0.$$

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According to the first three equations, $x_1, x_2, x_3$ are solutions of $$x^3 +px + q = 0$$

The fourth equation can be translated into $x_1x_2x_3 = x_3 + x_2$, i.e. $-q = -x_1$, then $x_1 = q$,

Conclude by noting that $q$ solves $x^3 +px + q = 0$

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