Working with Vieta

first of all, lets review Newtons sum $a+b+c = - \dfrac{3}{2},a^2 +b^2 +c^2 = \dfrac{45}{4}, a^3 +b^3 +c^3 = - \dfrac{117}{8},abc= 3$ now, simplify $\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=\dfrac{a^2 c-a^2 b+ab^2+b^2 c+c^2 b+ac^2-a^3 -b^3-c^3 -3abc}{-ac^2-a^2 b+a^2 c-cb^2+bc^2+ab^2 }$ and $\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}= \dfrac{b^2 c-bc^2+ac^2-a^2 c+a^2 b-ab^2}{abc}$ now notice that $\dfrac{b^2 c-bc^2+ac^2-a^2 c+a^2 b-ab^2}{-ac^2-a^2 b+a^2 c-cb^2+bc^2+ab^2}= -1$ , so the expression now is $\dfrac{a^2 c-a^2 b+ab^2+b^2 c+c^2 b+ac^2-a^3 -b^3-c^3 -3abc}{-abc}$ factorise $\dfrac{a^2 (b+c)+c^2 (a+b)+b^2(c+a)-(a^3 +b^3+c^3)-3abc}{-abc}$ from the first line we can tell $a+b=-\dfrac{3}{2}-c$ and so on, now substitute that to get $\dfrac{a^2 (-\dfrac{3}{2}-a)+c^2 (-\dfrac{3}{2}-c)+b^2(-\dfrac{3}{2}-b)-(a^3 +b^3+c^3)-3abc}{-abc}$ simplify and factorise again to get $\dfrac{-\dfrac{3}{2}(a^2 +b^2 +c^2)-2(a^3 +b^3+c^3)-3abc}{-abc}$ substitute the values by Newton's and we get the value is $\left| \dfrac{-9}{8}\right|$ and $9+8= \boxed{17}$

Before I give my solution, using Vieta's formula gives a + b + c = -3/2, ab + bc + ac = -9/2, and abc = 3.

Also the following identities are useful:

(a + b + c)^2 - 3(ab + bc + ac) = a^2 + b^2 + c^2 - ab - bc - ac = 63/4

a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac) + 3abc = (-3/2)(63/4) + 9 = -117/8.

(a + b + c)^3 = a^3 + b^3 + c^3 + 6abc + 3ab^2 + 3ba^2 + 3ac^2 + 3ca^2 + 3bc^2 + 3cb^2 or -27/8 = -117/8 + 18 + 3ab^2 + 3ba^2 + 3ac^2 + 3ca^2 + 3bc^2 + 3cb^2 which implies ab^2 + ba^2 + ac^2 + ca^2 + bc^2 + cb^2 = -9/4.

Now, the attack here is to make both factors one term which makes the first factor equal to 27/8 and the second -1/3 (after cancelling the denominator on the first and its negative on the second). This gives m/n = |-9/8| = 9/8. Hence, 17.