# How to express sum of three roots of a polynomial? [closed]

Let $\alpha _i,i=1,2,3$ be the roots of the polynomial

$x^{3}+px^{2}+qx+r$ where $p,q,r \in \mathbb{C}$ . Express the sum

$\sum_{i,j=1,i\neq j}^{3} \alpha _i^{2}\alpha_j$ in terms of $p, q$and $r$. I don't know how to proceed and get correct answer ,can u tell be book from which I can practice more such questions ?

• Start by explicitly writing out the given information in terms of mathematical formulae Commented Jul 14, 2017 at 20:53
• Did u try something? Commented Jul 14, 2017 at 20:56

Multiply $q = \alpha_1\alpha_2 + \alpha_1\alpha_3 + \alpha_2\alpha_3$ with $-p = \alpha_1 + \alpha_2 + \alpha_3$.
Your expression is equal to$$(\alpha_1+\alpha_2+\alpha_3)(\alpha_1\alpha_2+\alpha_2\alpha_3+\alpha_1\alpha_3)-3\alpha_1\alpha_2\alpha_3=-pq+3r.$$
Let's call the roots $a$, $b$ and $c$; the sum you want to compute is $$S(a,b,c)=a^2b+a^2c+ab^2+b^2c+ac^2+bc^2$$ We want to express this polynomial in $a,b,c$ as $$S(a,b,c)=A(a+b+c)^3+B(a+b+c)(ab+bc+ca)+Cabc$$ which is possible by the theory of symmetric polynomials.
If $a=1$, $b=0$ and $c=0$ we have $$S(1,0,0)=0=A$$ If $a=1$, $b=1$ and $c=0$ we have $$S(1,1,0)=2=8A+2B$$ If $a=b=c=1$ we have $$S(1,1,1)=6=27A+9B+C$$ Thus $A=0$, $B=1$ and $C=-3$. Therefore $$S(a,b,c)=(a+b+c)(ab+bc+ca)-3abc$$ Now apply Viète's formulas.
• The OP called $\alpha_1$, $\alpha_2$, and $\alpha_3$ the roots of the polynomial. Why did you decide to call them $a$, $b$, and $c$? Commented Jul 14, 2017 at 21:46