<s>[INST] Given that the quadratic equation $x^2-px+q=0$ has two roots $r$ and $s,$ Find the quadratic equation that takes $r^3$ and $s^3$ as its roots. 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?
 [/INST] 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
$$x^2 -p(p^2-3q)x +q^3=0$$
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