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- Feb 14, 2012

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- Thread starter anemone
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- #1

- Feb 14, 2012

- 3,802

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- Feb 14, 2012

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$a+b+c=-m\\ab+bc+ca=n\\abc=-p$

The given equality yields $n^3=m^3p$. Hence, if $m\ne 0$, the equation $P(x)=0$ can be written as

$x^3+mx^2+nx+\dfrac{n^3}{m^3}=0$

$m^3x^3+m^4x^2+nm^3x+n^3=0$

Factor the left-hand side,

$(mx+n)(m^2x^2-mnx+n^2)+m^3x(mx+n)=(mx+n)(m^2x^2+(m^3-mn)x+n^2)$

It follows that one of the roots of $P$ is $x_1=-\dfrac{n}{m}$ and the other two satisfy the condition $x_2x_3=\dfrac{n^2}{m^2}$. We obtained $x_1^2=x_2x_3$, thus the roots are terms of a geometric sequence.

If $m=0$ then $n=0$ but in this case, the polynomial $x^3+p$ cannot have three real roots.