Question

If p/(b-c)(b+c-2a)=q/(c-a)(c+a-2b)=r/(a-b)(a+b-2c) then find the value of p+q+r=?

Answer 1

First of all
$\frac{p}{(b - c)(b + c - 2a)} = \frac{q}{(c - a)(c + a - 2b)} = \alpha$
$\frac{r}{(a - b)(a + b - 2c)} = \alpha$

Then
$p = \alpha (b - c)(b + c - 2a)$
$q = \alpha (c - a)(c + a - 2b)$
$r = \alpha (a - b)(a + b - 2c)$
So on adding p, q, and r
We will get
0