MHB Show that p² + p - 6 is equal to q or r.

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The discussion focuses on proving that for the roots \( p, q, r \) of the polynomial \( x^3 - 9x + 9 = 0 \), the expression \( p^2 + p - 6 \) is equal to either \( q \) or \( r \). It is established that demonstrating \( p^2 + p - 6 \) as a root of the cubic polynomial, distinct from \( p \), suffices for the proof. The calculation shows that substituting \( p \) into the polynomial results in \( (p^2 + p - 6)^3 - 9(p^2 + p - 6) + 9 = 0 \). Since neither \( \pm \sqrt{6} \) is a root of the cubic, it follows that \( p^2 + p - 6 \) must equal either \( q \) or \( r \). This confirms the relationship between the roots and the expression.
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Let $p$, $q$ and $r$ be roots of polynomial $$x^3-9x+9=0$$. Show that $$p^2+p-6$$ is equal to $q$ or $r$.
 
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anemone said:
Let $p$, $q$ and $r$ be roots of polynomial $$x^3-9x+9=0$$. Show that $$p^2+p-6$$ is equal to $q$ or $r$.

It is sufficient to show that $$p^2+p-6$$ is a root of $$x^3-9x+9$$ different from $p$. We may as well also note that $$x^3-9x+9$$ 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 $$x^3-9x+9$$ 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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