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Here's what I've tried:

Using Vieta's formulas: $\;rs = q\;$ and $\,r+s = p$.

Then I cubed $rs$ which is $q^3$ and $(r+s)^3 + rs$ which is $p^3 + pq.$ Thinking that $x^2 - (p^3+pq)x + q^3$ is the answer. Did I do something wrong here?

dxiv
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Isaiah Leobrera
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  • Why did you take $(r+s)^3+rs$? Note that if your new quadratic equation is $x^2-ax+b$, then (using the Vieta formulas again) $b=r^3s^3$ (which you correlated well) but $a=r^3+s^3$, which doesn't correspond to anything that you suggested. – Steven Stadnicki Sep 08 '18 at 04:10
  • and (r+s)^3+rs You may want to elaborate this step some more (and realize why it's wrong). Rather, think at using $,(a+b)^3=a^3+b^3+3ab(a+b),$. – dxiv Sep 08 '18 at 04:12

2 Answers2

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The equation in question would look like $$(x-r^3)(x-s^3)=x^2-(r^3+s^3)x+r^3s^3=0$$ As you've already identified, the last term is $(rs)^3=q^3$.

The coefficient of $-x$ is $$\begin{split} r^3+s^3 &= (r+s)(r^2-rs+s^2) \\ &= p(r^2+2rs+s^2-3rs) \\ &= p\bigg((r+s)^2-3rs\bigg) \\ &= p(p^2-3q) \end{split}$$ Therefore, the equation is $$x^2 -p(p^2-3q)x +q^3=0$$

Stefan Lafon
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Use the following identities

$$r^3+s^3=(r+s)^3-3rs(r+s)$$

$$r^3s^3=(rs)^3$$

You know that $r+s=p$ and $rs=q$ by Vieta's formulas. Hence, the desired polynomial is given by

$$p(x)=x^2-(p^3-3qp)x+q^3$$

stressed out
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