Find new polynomial coefficients using Vieta's formulas 
$p(x)=x^3+ax^2+bx+c$ 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?
 A: 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*}
A: 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 $q(x)=x^3+2ax^2+(b+a^2)x+ab-c$.
A: 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:
$$0=p(x)=p(-a-y)=-(y+a)^3+a(y+a)^2-b(y+a)+c=-y^3-2ay^2-(a^2+b)y+c$$
