Question: > 
> 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?

Answer: 
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 $B$.
> 
> 
>
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 $24, -3$.