Question

If x/(b-c)(b+c-2a)=y/(c-a)(c+a-2b)=z/(a-b)(a+b-2c) then the value of x+y+z is

Answer 1

let (x/(b-c)(b+c-2a) = y/(c-a)(c+a-2b) = z/(a-b)(a+b-2c) = k

Re arranging,
x= k (b-c)(b+c-2a)
y= k (c-a)(c+a-2b)
z= k (a-b)(a+b-2c)

now x+y+z = k [ (b-c)(b+c-2a) + (c-a)(c+a-2b) + (a-b)(a+b-2c) ]
so x+y+z= k[ b.b+ bc-2ab-bc-c.c+2ac+c.c+ac-2bc-ac-a.a+2ab+ a.a +a.b-2ac-ab-b.b+2bc]
x+y+z = k * 0 = 0 answer