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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 ?

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

3 Answers 3

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Hint:

Multiply $q = \alpha_1\alpha_2 + \alpha_1\alpha_3 + \alpha_2\alpha_3$ with $-p = \alpha_1 + \alpha_2 + \alpha_3$.

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Your expression is equal to$$(\alpha_1+\alpha_2+\alpha_3)(\alpha_1\alpha_2+\alpha_2\alpha_3+\alpha_1\alpha_3)-3\alpha_1\alpha_2\alpha_3=-pq+3r.$$

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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.

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  • $\begingroup$ 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 $c$? $\endgroup$ Commented Jul 14, 2017 at 21:46
  • $\begingroup$ @JoséCarlosSantos Because they're easier to type and to read. $\endgroup$
    – egreg
    Commented Jul 14, 2017 at 22:01

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