The roots of the equation a(b–c) x² + b(c–a)x + c(a –b) = 0 are

The given equation is $a\left(b-c\right)x^2+b\left(c-a\right)x+c\left(a-b\right)=0$ 

We know that if the sum of the coefficients of quadratic equation $a{x}^2+bx+c=0$ is zero then the roots are $1,\frac{c}{a}$

In the given equation, the sum of the roots is $ab-ac+bc-ba+ca-cb=0$

hence one root is $1$ other  root is $\frac{c\left(a-b\right)}{a\left(b-c\right)}$