# 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) &= (1-2\alpha) + (1-2\beta) + (1-2\gamma) \\ &= 3 -2(\alpha+\beta+\gamma) = 3 +6 =9. \end{align}
Any symmetric (polynomial) function of the roots can be expressed in terms of the Vieta coefficients. Here, check the hint: $$\sum \alpha^2(\beta+\gamma) = (\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha)-3\alpha\beta\gamma$$
$$a,b,c$$ 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}
• Some explanation of some of the workings here might help too - eg whzat identity you are using at each step. I'm having trouble seeing how you get from line 2 to 3 in your solution. I can see $(a+b+c)$ being replaced by $-3$ but I don't understand why $a^3+b^3+c^3$ becomes $(a+b+c)^3$... For a complete answer you should also probably explain things like why $a+b+c = -3$ – Chris Feb 21 '19 at 13:11