If $a$, $b$, $c$ are the roots of $x^3-6x^2+3x+1=0$, find all possible values of $a^2b+b^2c+c^2a$ 
Let $a$, $b$, $c$ be the roots of
$$x^3 - 6x^2 + 3x + 1 = 0$$ Find all possible values of $a^2 b + b^2 c + c^2 a$. Express all the possible values, in commas.

I've already tried to bash out all the roots, Vieta's Formula, and try to manipulate the equation to become easier to work with. However, Vieta's didn't get me anywhere, and I couldn't find a way to make the equation simpler or anything. Any hints to start this problem?
 A: Hint : Let $A=a^2 b + b^2 c + c^2 a$ and $B=a^2 c + b^2 a + c^2 b$. Now calculate $A+B$ and $AB$.

 Now consider the quadratic whose roots are $A$ and $B$.

A: Let $p = a + b + c$, $q = ab + bc + ca$ and $r = abc$.
As pointed out by @Donald Splutterwit, we may let $A = a^2b + b^2c + c^2a$ and $B = ab^2 + bc^2 + ca^2$,
then $A + B$ and $AB$ are both symmetric which can both be expressed in terms of polynomials of $p, q, r$.
Indeed, we have
\begin{align} 
A + B &= a^2b + b^2c + c^2a + ab^2 + bc^2 + ca^2\\
&= a^2(b+c) + b^2(c+a) + c^2(a+b)\\
&= (a^2+b^2+c^2)(a+b+c) - (a^3 + b^3 + c^3)\\
&= (p^2 - 2q)p - [3r + p(p^2 - 2q - q)]\\
&= pq - 3r
\end{align}
where we have used the known identity
$$a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab - bc - ca).$$
Also, we have
\begin{align}
AB &= (a^3+b^3+c^3)abc + (a^3b^3 + b^3c^3 + c^3a^3) + 3(abc)^2\\
&= [3r + p(p^2 - 2q - q)]r + [q^3 - 6r^2 - 3r(pq - 3r)] + 3r^2\\
&= p^3r-6pqr+q^3+9r^2
\end{align}
where we have used
\begin{align}
(ab+bc+ca)^3 &= a^3b^3 + b^3c^3 + c^3a^3 + 6(abc)^2\\
&\quad + 3abc(a^2b + b^2c + c^2a + ab^2 + bc^2 + ca^2).
\end{align}
Now, by Vieta's formula, we have $p = 6, q = 3, r = -1$.
Thus, $A + B = 21$ and $AB = -72$ which results in $A = 24, B = -3$ or $A = -3, B = 24$.
Thus, all the possible values of $A$ are $24, -3$.
