If $\ x^3+px-q=0 $ has three roots $a$, $b$, $c$ then find an equation with roots $a+b$, $b+c$ and $c+a$

Question

The roots are not $a$, $b$, $c$ but $\alpha$, $\beta$, and $\gamma$. I wrote $a$, $b$, $c$, due to space constraints.

I know that I have to start with $$\ \alpha + \beta + \gamma = 0 $$ $$\ \alpha + \beta = -\gamma, \beta + \gamma = -\alpha, \alpha + \gamma = -\beta $$ but I don't know how to proceed hereafter.

Answer 1

As you have already noticed that

$$ \alpha +\beta =-\gamma$$

$$ \beta +\gamma =-\alpha$$ $$\alpha +\gamma=-\beta$$

This means that the new equation must have roots $-\alpha,-\beta,-\gamma$.

This means that if $x$ is the root of $f(x)$, then $-x$ is the root of the new equation.

$$f(-x)=-x^3-px-q$$

and the new equation is $$x^3+px+q=0$$

Answer 2

Well, you rather need to compute $$\begin{align}(a+b)+(b+c)+(a+c)&=2(a+b+c)\\&=0\\(a+b)(b+c)+(a+b)(c+a)+(b+c)(c+a)&=a^2 + 3 a b + 3 a c + b^2 + 3 b c + c^2\\&=(a+b+c)^2+(ab+bc+ca)\\&=p\\(a+b)(b+c)(c+a)&=a^2 b + a^2 c + a b^2 + 2 a b c + a c^2 + b^2 c + b c^2\\&=(ab+bc+ca)(a+b+c)-abc\\&=q\end{align}$$

which, by Vieta's formulas give you the coefficients of $$(x-(a+b))(x-(b+c))(x-(c+a))$$