Let `omega` be a complex cube root unity with `omega != 1.` A fair die is thrown three times. If `r_1,r_2 and r_3` are the numbers obtained on the die, then the probability that `omega^(r1) + omega^(r2) + omega^(r3)=0` is (a) `1/18` (b) `1/9` (c) `2/9` (d) `1/36`

Options

A. `1//18`
B. `1//9`
C. `2//9`
D. `1//36`

Explanation

To solve the problem, we need to find the probability that the sum \( \omega^{r_1} + \omega^{r_2} + \omega^{r_3} = 0 \) when a fair die is thrown three times, where \( \omega \) is a complex cube root of unity (specifically, \( \omega = e^{2\pi i / 3} \)) and \( \omega \neq 1 \).

### Step-by-Step Solution:

1. **Understanding Cube Roots of Unity**:
The complex cube roots of unity are \( 1, \omega, \) and \( \omega^2 \), where:
\[
\omega = e^{2\pi i / 3}, \quad \omega^2 = e^{4\pi i / 3}
...