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- Feb 14, 2012

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- Thread starter anemone
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- Thread starter
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- #1

- Feb 14, 2012

- 3,918

It is sufficient to show that \(\displaystyle p^2+p-6\) is a root of \(\displaystyle x^3-9x+9\) different from $p$. We may as well also note that \(\displaystyle x^3-9x+9\) has three distinct real roots.

Well substitute the first into the second and we get:

$$(p^2+p-6)^3-9(p^2+p-6)+9=(p^3-9p+9)(p^3+3p^2-6p-17)$$

but as $p$ is a root of \(\displaystyle x^3-9x+9\) we have:

$$(p^2+p-6)^3-9(p^2+p-6)+9=(p^3-9p+9)(p^3+3p^2-6p-17)=0$$

That is $p^2+p-6$ is a root of the cubic, and as neither $\pm \sqrt{6}$ is a root of the cubic $p^2+p-6 \ne p$ hence $p^2+p-6$ must be equal to $q$ or $r$.

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