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

Question

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

A) $p^3 -8r$

B) $4pq-p^3$

C) $4pq-p^3-8r$

D) $4pq-8r$

Solution:

$$a+b+c= p$$

$$ab+bc+ca= q$$

$$abc= r$$

Using above we get:

$$(p-2a)(p-2b)(p-2c)$$

Answer 1

HINT:

$a+b-c=a+b+c-2c=p-2c=y\iff c=\dfrac{p-y}2 $

As $c$ is a root of the given equation, set the value of $c$ to form a cubic equation in $y$ on simplification.

Finally use Vieta's formula