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.

Question

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?

Answer 1

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

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

Answer 2

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 $$x^2 -p(p^2-3q)x +q^3=0$$