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 $ab+bc+ac=0$.
 A: Let
$$ z_1 = a + b + c \\
z_2 = a + \epsilon b + \epsilon^2 c \\\
z_3 = a + \epsilon^2 b + \epsilon c $$
Observe that $z_1 \in \Bbb{R}$, and $z_2 = \overline {z_3}$, so there is some $r$ and $\theta$ such that $z_2 = re^{i \theta}$ and $z_3 = re^{-i\theta}$.
$$r^2 = z_2 z_3 = (a+\epsilon b+\epsilon^2 c)(a + \epsilon^2 b+\epsilon c) = a^2 + b^2 + c^2 + \epsilon (ab + ac + bc) + \epsilon^2 (ab + ac + bc)$$
Plugging in $ab + bc + ac = 0$ twice we get
$$r^2 = a^2 + b^2 + c^2$$
Adding twice again $ab + bc + ac = 0$, we get
$$r^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac = (a+b+c)^2$$
leading to the conclusion
$$|a+b+c| = r = |z_2| =  |z_3| = |z_1| < 1$$
which means the limit must be 0.
A: 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 $0$.
