if $a,b,c$ are the roots of the equation $x^3-px^2+qx-r=0$ find the value of $(b+c)(c+a)(a+b)$

Question

Using vieta's formula:

$a+b+c=p$

$ab + ac +bc = q$

$abc = r$

on expansion of the bracket,

$(b+c)(c+a)(a+b) $

$=(bc+c^2+ab+ac)(a+b)$ $=(q+c^2)(a+b)$ after this step I am unable to proceed further. Could someone please help me out?

Welcome to MSE. A question should be written in such a way that it can be understood even by someone who did not read its title. — José Carlos Santos, Mar 10 '24 at 11:15
Hint: the expression becomes $(p-a)(p-b)(p-c)$ which is just the cubic evaluated at $p$. — Sahaj, Mar 10 '24 at 11:18
@Sahaj Thank you, that solves the problem. After factorization it becomes $-r + p(ab+ac+bc)-p^2(a+b+c)+p^3$ — littlesniper23, Mar 10 '24 at 11:32

score 3 · Answer 1 · answered Mar 10 '24 at 11:24

3

Since $a,b,c$ is the solutions, we have $$(x-a)(x-b)(x-c)=0=x^3-px^2+qx-r,$$ and let $x=a+b+c$, we have $$(b+c)(c+a)(a+b)=(a+b+c-a)(a+b+c-b)(a+b+c-c)\\=(a+b+c)^3-p(a+b+c)^2+q(a+b+c)-r$$

answered Mar 10 '24 at 11:24

Perry_W

85

Well, you may want to set $x=p$ instead, to get $pq-r$. – Macavity Mar 10 '24 at 12:20

if $a,b,c$ are the roots of the equation $x^3-px^2+qx-r=0$ find the value of $(b+c)(c+a)(a+b)$

1 Answers1