The two roots of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ are $1$ and: 
$\textbf{(A)}\ \frac{b(c-a)}{a(b-c)}\qquad\textbf{(B)}\ \frac{a(b-c)}{c(a-b)}\qquad\textbf{(C)}\ \frac{a(b-c)}{b(c-a)}\qquad\textbf{(D)}\ \frac{c(a-b)}{a(b-c)}\qquad\textbf{(E)}\ \frac{c(a-b)}{b(c-a)}$
Let the other root be $k$. Then by Vieta's Formulas, $k\cdot 1=\frac{c(a-b)}{a(b-c)}$, $\fbox{D}$