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Let a,b,ca,b,ca,b,c be the real roots of the cubic equation 2t3+3t2−9t−6=02t^3+3t^2-9t-6=02t3+3t2−9t−6=0. Suppose that ∣(ab−c+bc−a+ca−b)(b−ca+c−ab+a−bc)∣=mn\left|\left(\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}\right)\left(\dfrac{b-c}{a}+\dfrac{c-a}{b}+\dfrac{a-b}{c}\right)\right|=\dfrac{m}{n}∣∣∣∣(b−ca+c−ab+a−bc)(ab−c+bc−a+ca−b)∣∣∣∣=nm for some relatively prime positive integers mmm and nnn. Find m+nm+nm+n.
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