Newtons polynomial

We solve this using Vieta's formulas. Let $P_n = a^n+b^n+c^n$ , $S_1 = a+b+c = 0$ , $S_2 = ab+bc+ca = M$ and $S_3 = abc=N$ . Then, we have:

\(\begin{array} {} P_2 = a^2 + b^2+c^2 = S_1P_1 - 2S_2 = (0)(0)-2M & = -2M \\ P_3 = a^3 + b^3+c^3 = 0-M(0)+3N & = 3N \\ P_4 = a^4 + b^4+c^4 = 0-M(-2M)+N(0) & = 2M^2 \\ P_5 = a^5 + b^5+c^5 = 0-M(3N)+N(-2M) = -5M = 90 & \Rightarrow MN = - 18 & ...(1) \\ P_6 = a^6 + b^6+c^6 = 0 -M(2M^2) + N(3N) = 227 & \Rightarrow -2M^3 + 3N^2 = 227 &...(2) \end{array} \)

Solving using equations (1) and (2), we found that $M=2$ and $N=-9$ .

Therefore, $M^5+N^5 = 2^5 + (-9)^5 = 32-59049 = \boxed{-59017}$