Since this is rotation-dilation symmetric, w.l.o.g. $b=0, a=1$. Then letting $x=-c$ the equation $1+x^n+(-x-1)^n=0$ is only allowed to have the solutions $x=\omega$ and $x=\omega^2$ where $\omega$ is a third root of unity. If $n$ is odd, it certainly also has the solution $x=0$. So $n$ must be even and our equation is $(x+1)^n+x^n+1=0$. In particular, this is a polynomial of degree $n$ and hence must be a product of $n$ linear factors. Since this is a real polynomial, complex roots come in pairs of complex conjugates so that $\omega$ and $\omega^2$ must occur with the same multiplicity so that we must have the identity $(x+1)^{2k}+x^{2k}+1=2(x^2+x+1)^k$ where $n=2k$. But now comparing coefficients of $x^{2k-3}$ on both sides we have $\binom{2k}{3}=2\binom{k}{3}+4\binom{k}{2}$ and hence $k=2$. So the only possible solution is $n=4$ and it's easy to check that $(x+1)^4+x^4+1=2(x^2+x+1)^2$ so that this is indeed a solution.

@above Tintarn wrote: If $n$ is odd, it certainly also has the solution $x=0$. So $n$ must be even Yes, it also has the solution $x=0$ when $n$ is odd. But the condition says that CatalinBordea wrote: Find all natural numbers $ n\ge 4 $ that satisfy the property that the affixes of any nonzero pairwise distinct complex numbers $ a,b,c $ that verify the equation $$ (a-b)^n+(b-c)^n+(c-a)^n=0, $$ I think that we need to choose three pairwise different $a,b,c$, and then $a,b,c$ satisfy $ (a-b)^n+(b-c)^n+(c-a)^n=0$, not $a,b,c$ should be different and satisfy $ (a-b)^n+(b-c)^n+(c-a)^n=0$ So $n$ is odd is not all wrong in that comprehension And we can get the solution of $n=4,5,7$

Yes, you are right. Indeed, $x=0$ and $x=-1$ are also allowed. For even $n$, this changes nothing since they are clearly not solutions of our equation. For odd $n=2k+1$, we get that all the solutions of $(x+1)^n=x^n+1$ must be in $\{0,-1,\omega,\omega^2\}$. However, clearly $x=0$ and $x=-1$ are solutions of multiplicity exactly $1$, so comparing coefficients of $x^{2k}$ we must have \[(x+1)^{2k+1}-x^{2k+1}-1=(2k+1)x(x+1)(x^2+x+1)^{k-1}.\]Comparing coefficients of $x^{2k-2}$ we get \[\binom{2k+1}{3}=(2k+1) \cdot \left(k-1+\binom{k}{2}\right)\]and hence $k=2$ or $k=3$ and hence $n=5$ or $n=7$ which indeed both work.