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?

Let $t$ be the $3$rd root of the equation $x^3 - cx + d = 0$
So $$t^3 - ct + d = 0 \tag{1}$$
Also $$r + s + t  = 0 \implies t = -(r+s) \tag{2}$$
Substitute the value of $t$ from $(2)$ into $(1)$ to get
$$-(r+s)^3 + c(r+s) + d = 0 \equiv (r+s)^3 - c(r+s) - d = 0$$
Hence the proof.
