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.
 A: 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)
&= (1-2\alpha) + (1-2\beta) + (1-2\gamma) \\
&= 3 -2(\alpha+\beta+\gamma) = 3 +6 =9.
\end{align}
A: Any symmetric (polynomial) function of the roots can be expressed in terms of the Vieta coefficients.  Here, check the hint:
$$\sum \alpha^2(\beta+\gamma) = (\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha)-3\alpha\beta\gamma$$
--
In case you want a systematic method to express in terms of elementary symmetric polynomials, check this answer for Gauss' algorithm.
A: $a,b,c$ are the three roots.
$$
\begin{align}
&a^2*(b+c)+b^2*(a+c)+c^2*(a+b)\\
={}&(a+b+c)*(a^2+b^2+c^2)-(a^3+b^3+c^3)\\
={}&(-3)*(a^2+b^2+c^2)-(a+b+c)^3\\
={}&(-3)*((a+b+c)^2-2ab-2ac-2bc)-(a+b+c) * (a^2+b^2+c^2) + ab(a+b) + ac(a+c) + bc(b+c)\\
={}& (-3)*(9-2*(-2))-(-3)*(9-2*(-2)) + ab(a+b+c-c) + ac(a+b+c-b) + bc(a+b+c-a)\\
={}&(a+b+c)(ab+ac+bc)-3abc\\
={}&(-3)*(-2)-3*(-1)\\
={}&6-(-3)\\
={}&9
\end{align}
$$
