Question 25The two roots of the equation $$ \begin{aligned} a(b-c)x^2+b(c-a)x\\ {}+c(a-b)&=0 \end{aligned} $$ are $1$ and:$\dfrac{b(c-a)}{a(b-c)}$$\dfrac{a(b-c)}{c(a-b)}$$\dfrac{a(b-c)}{b(c-a)}$$\dfrac{c(a-b)}{a(b-c)}$$\dfrac{c(a-b)}{b(c-a)}$