Let $x=-(ab+bc+ca)$ and WLOG (by scaling) assume that $abc=1$. Then in view of Newton sums, let $P_m(x)$ denote the polynomial representation of $a^m+b^m+c^m$ in terms of the elementary symmetric polynomials $a+b+c=0$, $ab+bc+ca=-x$, and $abc=1$ so that $P_0(x)=3$, $P_1(x)=0$, $P_2(x)=2x$, and $P_m(x)=xP_{m-2}(x)+P_{m-3}(x)$ for $m\ge3$. Now consider the roots of $P_n(x)$ for $n>1$.

First get rid of silly $abc=0$ cases and stuff. Now if (by scaling) $abc = 1$ and $x = -(ab + bc + ca)$, then $P_n(x) = xP_ {n-2}(x) + P_ {n-3}(x)$ with initial conditions $P_0=3$, $P_1=0$, $P_2=2x$. We want to show $P_n (x)$ has real roots times a third root of unity, modulo trivial cases. We first take out the $0$ roots and do $x^3\to x$ to get new polynomials satisfying $Q_n=xQ_{n-2}+Q_{n-3}$ whenever $3\mid n$ and $Q_n=Q_{n-2}+Q_{n-3}$ otherwise (starting with $3,0,2$). We induct in groups of six to show that for all $m\ge1$, the degrees for $Q_{6m},Q_{6m+2},\ldots,Q_{6m+5}$ are $m$ and $Q_{6m+1}$ has degree $m-1$, and ordering the roots from least to greatest (coefficients are all positive, so roots are clearly all negative), we have \begin{align*} -\infty<r_{6m+2,1} &<r_{6m,1}<r_{6m+1,1}<r_{6m+2,2} \\ &<\cdots<r_{6m+1,m-1}<r_{6m+2,m}<r_{6m,m}<0 \end{align*}and \begin{align*} -\infty<r_{6m+4,1} &<r_{6m+5,1}<r_{6m+3,1}<r_{6m+4,2} \\ &<\cdots<r_{6m+3,m-1}<r_{6m+4,m}<r_{6m+5,m}<r_{6m+3,m}<0. \end{align*}For the first induction case, use IVT first for $6m+3,6m+4$ to get \begin{align*} -\infty &< r_ {6 m + 4, 1} < r_ {6 m + 2, 1} < r_ {6 m, 1} < r_ {6 m + 3, 1} < r_ {6 m + 1, 1} < r_ {6 m + 4, 2} < r_ {6 m + 2, 2} \\ &< \cdots < r_ {6 m + 1, m - 1} < r_ {6 m + 4, m} < r_ {6 m + 2, m} < r_ {6 m, m} < r_ {6 m + 3, m} < 0. \end{align*}Now we use $Q_ {6 m + 5} = Q_ {6 m + 3} + Q_ {6 m + 2}$ to get $r_ {6 m + 2, i} < r_ {6 m + 5, i} < r_ {6 m + 3, i}$ for all $i$, so \begin{align*} -\infty &< r_ {6 m + 4, 1} < r_ {6 m + 2, 1} < r_ {6 m + 5, 1} < r_ {6 m + 3, 1} < r_ {6 m + 1, 1} < r_ {6 m + 4, 2} < r_ {6 m + 2, 2} \\ &< \cdots < r_ {6 m + 1, m - 1} < r_ {6 m + 4, m} < r_ {6 m + 2, m} < r_ {6 m + 5, m} < r_ {6 m + 3, m} < 0. \end{align*}For the second induction case, use IVT first for $6m+6,6m+7$ to get \begin{align*} -\infty &< r_ {6 m + 6, 1} < r_ {6 m + 4, 1} < r_ {6 m + 7, 1} < r_ {6 m + 5, 1} < r_ {6 m + 3, 1} < r_ {6 m + 6, 2} < r_ {6 m + 4, 2} \\ &< \cdots < r_ {6 m + 3, m - 1} < r_ {6 m + 6, m} < r_ {6 m + 4, m} < r_ {6 m + 7, m} < r_ {6 m + 5, m} < r_ {6 m + 3, m} < r_ {6 m + 6, m + 1} < 0. \end{align*}Now we use $Q_ {6 m + 8} = Q_ {6 m + 6} + Q_ {6 m + 5} $ to get \begin{align*} -\infty &< r_ {6 m + 8, 1} < r_ {6 m + 6, 1} < r_ {6 m + 4, 1} < r_ {6 m + 7, 1} < r_ {6 m + 5, 1} < r_ {6 m + 8, 2} < r_ {6 m + 6, 2} < r_ {6 m + 4, 2} \\ &< \cdots < r_ {6 m + 6, m} < r_ {6 m + 4, m} < r_ {6 m + 7, m} < r_ {6 m + 5, m} < r_ {6 m + 8, m + 1} < r_ {6 m + 6, m + 1} < 0, \end{align*}completing the induction.