If a root of the given equation
a(b-c)x2+b(c-a)x+c(a-b)=0 is 1, then the other will be
a(b-c)b(c-a)
b(c-a)a(b-c)
c(a-b)a(b-c)
None of these
Let α=1 and the other root is β, then product of roots 1.β=c(a-b)a(b-c)⇒β=c(a-b)a(b-c)