$p(x)=x^3+ax^2+bx+c$ has roots $x_1,x_2,x_3$. Find $a,b,c$ for $q(x)$ which has roots at $x_1+x_2$, $x_1+x_3$ and $x_2+x_3$.
I know that I'm supposed to use Vieta's formulas where $$x_1+x_2+x_3=-a$$ $$x_1x_2+x_1x_3+x_2x_3=b$$ $$x_1x_2x_3=-c$$ but I do not know how to tackle this problem. Any solutions or tips?
Given that $x_1,x_2,x_3$ are the roots of the polynomial $$x^3+ax^2+bx+c$$ then by Vieta's formulas $$x_1+x_2+x_3=-a,\;x_1x_2+x_2x_3+x_1x_3=b,\;x_1x_2x_3=-c.$$ Again by Vieta's formulas the polynomial with roots $x_1+x_2$, $x_2+x_3$, $x_3+x_1$ is equal to $$x^3+Ax^2+Bx+C$$ where \begin{align*} -A&=(x_1+x_2)+(x_2+x_3)+(x_3+x_1)=2(x_1+x_2+x_3)=-2a\\ B&=(x_1+x_2)(x_2+x_3)+(x_2+x_3)(x_3+x_1)+(x_3+x_1)(x_1+x_2)=a^2+b\\ -C&=(x_1+x_2)(x_2+x_3)(x_3+x_1)=c-ab. \end{align*}
First I want to mention that you should obtain that $a$ is the component of your second order term, hier $a$ should be replaced by the third order term, which is $1$, so you should obtain that \begin{align} &x_1+x_2+x_3=-a,\\ &x_1x_2+x_2x_3+x_3x_1=b,\\ &x_1x_2x_3=-c. \end{align} Now let us suppose that $q(x)=x^3+a'x^2+b'x+c'$, which has roots $x_1+x_2,x_2+x_3,x_3+x_1$. Then again from Vieta's formula \begin{align} -a'&=2(x_1+x_2+x_3)=-2a,\\ b'&=(x_1+x_2)(x_2+x_3)+(x_2+x_3)(x_3+x_1)+(x_3+x_1)(x_1+x_2)\\ &=(x_1x_2+x_2x_3+x_3x_1)+(x_1+x_2+x_3)^2,\\ &=b+a^2,\\ -c'&=(x_1+x_2)(x_2+x_3)(x_3+x_1)\\ &=-(a+x_1)(a+x_2)(a+x_3)\\ &=-(a^3+(x_1+x_2+x_3)a^2+(x_1x_2+x_2x_3+x_3x_1)a+x_1x_2x_3)\\ &=c-ab. \end{align} So the new polynimil is $q(x)=x^3+2ax^2+(b+a^2)x+ab-c$.
Since $x_1+x_2+x_3=-a$ by Vieta's formulas, the new roots can be written as: $$y_1=x_2+x_3=-a-x_1\,,\quad y_2=-a-x_2\,,\quad y_3=-a-x_3\,$$
With the subsitution $y=-a-x \iff x = -y-a$ the equation satisfied by $y_{1,2,3}$ is:
$$0=p(x)=p(-a-y)=-(y+a)^3+a(y+a)^2-b(y+a)+c=-y^3-2ay^2-(a^2+b)y+c$$
