Roots of the equation $a(b-c) x^2+b(c-a) x+c(a-b)$ $=0$ are
Roots of the equation $a(b-c) x^2+b(c-a) x+c(a-b)$ $=0$ are
$\frac{a(b-c)}{c(a-b)}, 1$
$\frac{b(c-a)}{c(a-b)}, 1$
$\frac{c(a-b)}{a(b-c)}, 1$
$\frac{c(a-b)}{b(c-a)}, 1$
Solution
$a(b-c) x^2+b(c-a) x+c(a-b)=0$
$x=1$ satisfies his equation
Product of roots $=\frac{c(a-b)}{a(b-c)}=1 \times \alpha$
$\alpha=\frac{c(a-b)}{a(b-c)}=\text { other root }$ Roots are $\frac{c(a-b)}{a(b-c)}$ and 1