Calculate $\lim_{n\to\infty} ((a+b+c)^n+(a+\epsilon b+\epsilon^2c)^n+(a+\epsilon^2b+\epsilon c)^n)$ Calculate $\lim_{n\to\infty} ((a+b+c)^n+(a+\epsilon b+\epsilon^2c)^n+(a+\epsilon^2b+\epsilon c)^n)$ with $a,b,c \in \Bbb R$ and $\epsilon \in \Bbb C \setminus \Bbb R, \epsilon^3=1$.
Since $a+\epsilon b + \epsilon^2c=\overline {a+\epsilon^2b+\epsilon c}$, the expression above should be real, but I don't know how to deal with the $\epsilon$ in order to bring it to a form where I can calculate its limit. 
Also, for this problem, $|a+b+c|<1$ and $ab+bc+ac=0$.

Here's a more brute force method.
The condition $ab+bc+ac=0$ says that the point $(a,b,c)$ lies on a cone with axis $a=b=c$.  Making the change of variables
$$
\left(
\begin{array}{c}
 a \\
 b \\
 c
\end{array}
\right) = \left(
\begin{array}{ccc}
 \frac{1}{6} \left(3+\sqrt{3}\right) & \frac{1}{6} \left(-3+\sqrt{3}\right) & \frac{1}{\sqrt{3}} \\
 \frac{1}{6} \left(-3+\sqrt{3}\right) & \frac{1}{6} \left(3+\sqrt{3}\right) & \frac{1}{\sqrt{3}} \\
 -\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}
\end{array}
\right) 
\left(
\begin{array}{c}
 u \\
 v \\
 w/\sqrt{2}
\end{array}
\right)
$$
(see rotation matrix from axis and angle) the conditions
$$
|a+b+c| < 1 \quad \text{and} \quad ab + bc + ac = 0
$$
become
$$
|w| < \sqrt{\frac{2}{3}} \quad \text{and} \quad u^2 + v^2 = w^2,
$$
and the main quantity in question becomes
$$
\left(\sqrt{\frac{3}{2}}\,w\right)^n + \left[\frac{\sqrt{3}}{4}-\frac{3}{4} + i \left(\frac{\sqrt{3}}{4}+\frac{3}{4}\right)\right]^n \left[ (v-iu)^n + e^{n i 4\pi/3}  (u-iv)^n \right].
\tag{1}
$$
Now since $|w| < \sqrt{2/3}$ and $u^2 + v^2 = w^2$ we know that $|v-iu| = |u-iv| < \sqrt{2/3}$, so, because
$$
\left|\frac{\sqrt{3}}{4}-\frac{3}{4} + i \left(\frac{\sqrt{3}}{4}+\frac{3}{4}\right)\right| = \sqrt{\frac{3}{2}},
$$
we may conclude that the absolute value of every term in $(1)$ is $<1$, and thus that the limit as $n\to\infty$ must be $0$.
