<s>[INST] Show that if $r$ and $s$ are positive roots to the equation $x^3-cx+d=0$, then $r+s$ is a root of $x^3-cx-d=0$ If $r$ and $s$ are roots of $x^3-cx+d=0$, then
$$r^3-cr+d=s^3-cs+d , $$
for $r \neq s$. Solving the equation for $c$ gives me
$$ c=r^2+rs+s^2$$
Now to find an expression for $d$
$$ r^3-cr+d=0 $$
$$ d=r(c-r^2) $$
Using the expression for $c$ gives me
$$d=sr^2+rs^2$$
Assume $t$ is a root of the second equation
$$t^3-ct+d=0$$
We can represent $c$ and $d$ in terms of $r$ and $s$ which gives us
$$s*r^2+r*s^2=t(t^2-r^2-rs-s^2)$$
Plugging this into wolfram alpha gives us $t=r+s$, which is what we wanted to show. But how can I solve for $t$ algebraically? Or perhaps there is another way to solve this question?
 [/INST] Let $p(x) = x^3-xc-d$ then $p(s)=p(r)=0$.

 This is the part you already made
 $$s^3-cs =d = r^3-rc$$ so
 $$ (s-r)(r^2+rs+s^2)= (s-r)c$$ so
 $$r^2+rs+s^2=c$$

Let $q(x)=x^3-cx+d$.
Now we have \begin{align}q(s+r)&=(s+r)^3-(r+s)c+d\\
&= p(s)+p(r)+3d +3sr(r+s)\\
&= 0+0+3d-3s(c-s^2)\\
&= 3p(s)\\
&=0
\end{align}
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