Do you know recurrence formula?

Let $P_{n}=a^n+b^n+c^n$ .

By Newton's Sums,

$(6)P_{1}-6=0\Rightarrow P_{1}=1 \\ (6)P_{2}-(6)P_{1}-6=0 \Rightarrow P_{2}=2 \\ (6)P_{3}-(6)P_{2}-(3)P_{1}-3=0 \Rightarrow P_{3}=3 \\ (6)P_{4}-(6)P_{3}- (3)P_{2}-P_{1}=0 \Rightarrow P_{4}=\frac{25}{6}$

We need to find, $\frac{P_{4}}{P_{3}}=\frac{\frac{25}{6}}{3}= \frac{25}{18} \Rightarrow 25+18=\boxed{43}$

Let $\displaystyle P_n = a^n + b^n + c^n$ for $n = 1, 2, 3, 4$ . Also, we let:

$\displaystyle S_1 = a + b + c = 1$ $\displaystyle S_2 = ab + bc + ca = -\frac{1}{2}$ $S_3 = abc = \frac{1}{6}$ with their corresponding value got from the Vieta's Identities . Now, by Newton's Sums , we have the following relationships between $P_n$ and $S_n$ : $\displaystyle P_1 = S_1$ $\displaystyle P_2 = S_1P_1 - 2S_2$ $\displaystyle P_3 = S_1P_2 - S_2P_1 + 3S_3$ $\displaystyle P_4 = S_1P_3 - S_2P_2 + S_3P_1$ Thus, we get $\displaystyle P_1 = 1$ $\displaystyle P_2 = 1 - 2(-\frac{1}{2}) = 2$ $\displaystyle P_3 = 2 + \frac{1}{2} + 3(\frac{1}{6}) = 3$ $\displaystyle P_4 = 3 + (\frac{1}{2})(2) + \frac{1}{6} = \frac{25}{6}$ The following values above gives $\displaystyle \frac{P_4}{P_3}$ = $\frac{m}{n}$ = $\displaystyle \frac{\frac{25}{6}}{3}$ = $\displaystyle \frac{25}{18}$ . Hence, the desired value of $\displaystyle m + n = 25 + 18 = 43$ .