Relation between roots and coefficients - manipulation of identities The polynomial $x^3+3x^2-2x+1$ has roots $\alpha, \beta, \gamma$ . Find $$\alpha^2(\beta + \gamma) + \beta^2(\alpha + \gamma) + \gamma^2(\alpha + \beta)$$


I tried finding the relation using $-b/a$, $c/a$ and $-d/a$. I couldn’t seem to find anything. I also tried solving for one root but it gave me back the polynomial but with the root as the variable. Also the polynomial can not be factorised.
Since the polynomial has three roots and its highest degree is 3, we can write
$$
p(x) = (x-\alpha)(x-\beta)(x-\gamma) = x^3 +3x^2 - 2x + 1.
$$
It then follows from
$$
x^3 - (\alpha+\beta+\gamma)x^2 + (\alpha\beta + \beta \gamma + \gamma \alpha)x - \alpha\beta \gamma = x^3+3x^2-2x + 1
$$
that
$$
\alpha+\beta+\gamma = -3, \quad \alpha\beta + \beta \gamma + \gamma \alpha = -2, \quad
\alpha\beta \gamma = -1.
$$
Note that 
$$
-2\alpha = \alpha(\alpha\beta + \beta \gamma + \gamma \alpha) = \alpha^2(\beta+\gamma) +\alpha\beta\gamma = \alpha^2(\beta+\gamma) - 1.
$$
Thus $\alpha^2(\beta + \gamma)=1-2\alpha$.
Similarly, $\beta^2(\alpha + \gamma) = 1-2\beta$ and $\gamma^2(\alpha + \beta) = 1-2\gamma$.
Therefore,
\begin{align}
\alpha^2(\beta + \gamma) + \beta^2(\alpha + \gamma) + \gamma^2(\alpha + \beta)
&amp;= (1-2\alpha) + (1-2\beta) + (1-2\gamma) \\
&amp;= 3 -2(\alpha+\beta+\gamma) = 3 +6 =9.
\end{align}
$a,b,c$ are the three roots.
$$
\begin{align}
&amp;a^2*(b+c)+b^2*(a+c)+c^2*(a+b)\\
={}&amp;(a+b+c)*(a^2+b^2+c^2)-(a^3+b^3+c^3)\\
={}&amp;(-3)*(a^2+b^2+c^2)-(a+b+c)^3\\
={}&amp;(-3)*((a+b+c)^2-2ab-2ac-2bc)-(a+b+c) * (a^2+b^2+c^2) + ab(a+b) + ac(a+c) + bc(b+c)\\
={}&amp; (-3)*(9-2*(-2))-(-3)*(9-2*(-2)) + ab(a+b+c-c) + ac(a+b+c-b) + bc(a+b+c-a)\\
={}&amp;(a+b+c)(ab+ac+bc)-3abc\\
={}&amp;(-3)*(-2)-3*(-1)\\
={}&amp;6-(-3)\\
={}&amp;9
\end{align}
$$