And counting.

Suppose that $a,b,c$ are the roots of the cubic equation $X^3 - 3X^2 + uX - v = 0$ . Then using Vieta's formulae, we deduce that $\begin{aligned} S_2 & = \; a^2 + b^2 + c^2 \; = \; 9 - 2u \\ S_3 & = \; a^3 + b^3 + c^3 \; = \; 27 - 9u + 3v \\ S_4 & = \; a^4 + b^4 + c^4 \; = \; 2u^2 - 36u + 12v + 81 \end{aligned}$ Since $S_3 = 15$ we deduce that $v = 3u-4$ , and hence $35 \; = \; S_3 \; = \; 2u^2 - 36u + 12(3u-4) + 81 \; = \; 2u^2 + 33$ so that $u^2 = 1$ . Since $S_2 = 9 - 2u < 10$ , we deduce that $u=1$ , so that $v=-1$ and $S_2 = 7$ . Thus $a,b,c$ are the roots of the cubic equation $X^3 - 3X^2 + X + 1 = 0$ , and hence Newton's formulae tell us that $S_5 \; = \; a^5 + b^5 + c^5 \; = \; 3S_4 - S_3 - S_2 \; = \; 3\times35 - 15 - 7 \; = \; \boxed{83}$