If `alpha=root(3)(1)` and `alpha` is not real then `alpha^(3n+1)+alpha^(3n+3)+alpha^(3n+5)` has the value

Options

A. -1
B. 0
C. 1
D. 3

Answer

To solve the problem, we need to evaluate the expression \( \alpha^{3n+1} + \alpha^{3n+3} + \alpha^{3n+5} \) where \( \alpha \) is a non-real cube root of unity. The cube roots of unity are \( 1, \omega, \omega^2 \), where \( \omega = e^{2\pi i / 3} \) and \( \omega^2 = e^{-2\pi i / 3} \). Since \( \alpha \) is not real, we will take \( \alpha = \omega \) or \( \alpha = \omega^2 \).

### Step-by-Step Solution:

1. **Identify the Value of Alpha**:
Since \( \alpha \) is a non-real cube root of unity, we can choose \( \alpha = \omega \) (where \( \omega = e^{2\pi i / 3} \)).

2. **Rewrite the Expression**:
The expression we want to evaluate is:
\[ \alpha^{3n+1} + \alpha^{3n+3} + \alpha^{3n+5} \]
Substituting \( \alpha = \omega \):
\[ \omega^{3n+1} + \omega^{3n+3} + \omega^{3n+5} \]

3. **Factor Out Common Terms**:
We can factor out \( \omega^{3n} \):
\[ \omega^{3n} (\omega^1 + \omega^3 + \omega^5) \]

4. **Simplify Using Properties of Cube Roots of Unity**:
We know that \( \omega^3 = 1 \). Thus:
- \( \omega^3 = 1 \)
- \( \omega^4 = \omega \)
- \( \omega^5 = \omega^2 \)

Therefore, we can rewrite the expression inside the parentheses:
\[ \omega^1 + \omega^3 + \omega^5 = \omega + 1 + \omega^2 \]

5. **Use the Sum of Cube Roots of Unity**:
We know from the property of cube roots of unity that:
\[ 1 + \omega + \omega^2 = 0 \]
Therefore:
\[ \omega + 1 + \omega^2 = 0 \]

6. **Final Evaluation**:
Now substituting back, we have:
\[ \omega^{3n} \cdot 0 = 0 \]

### Conclusion:
The value of the expression \( \alpha^{3n+1} + \alpha^{3n+3} + \alpha^{3n+5} \) is:
\[ \boxed{0} \]