If z is nonreal root of `[-1]^[1/7]` then,find the value of `z^86`+`z^175`+`z^289`

Explanation

To solve the problem, we need to find the value of \( z^{86} + z^{175} + z^{289} \) where \( z \) is a non-real root of \( (-1)^{1/7} \).

### Step-by-Step Solution:

1. **Identify the Non-Real Roots of \((-1)^{1/7}\)**:
The seventh roots of \(-1\) can be expressed in polar form. The principal root is \( e^{i\pi} \), and the seventh roots are given by:
\[ z_k = e^{i(\pi + 2k\pi)/7} \quad \text{for } k = 0, 1, 2, 3, 4, 5, 6 \]
The non-real roots correspond to \( k = 1, 2, 3, 4, 5, 6 \).

2. **Express \( z^{86}, z^{175}, z^{289} \)**:
Using the property of exponents, we can express \( z^{n} \) in terms of \( z^{7} \):
\[ z^7 = -1 \]
Hence, we can reduce the exponents modulo 7:
- For \( z^{86} \):
\[ 86 \mod 7 = 2 \quad \Rightarrow \quad z^{86} = z^2 \]
- For \( z^{175} \):
\[ 175 \mod 7 = 0 \quad \Rightarrow \quad z^{175} = z^0 = 1 \]
- For \( z^{289} \):
\[ 289 \mod 7 = 1 \quad \Rightarrow \quad z^{289} = z^1 = z \]

3. **Combine the Results**:
Now we can combine the results:
\[ z^{86} + z^{175} + z^{289} = z^2 + 1 + z \]

4. **Finding \( z^2 + z + 1 \)**:
To evaluate \( z^2 + z + 1 \), we can use the fact that \( z = e^{i(\pi + 2k\pi)/7} \). For any non-real root, we can consider \( z = e^{i\pi/7} \) (for \( k=0 \)):
\[ z^2 = e^{2i\pi/7}, \quad z = e^{i\pi/7} \]
Thus:
\[ z^2 + z + 1 = e^{2i\pi/7} + e^{i\pi/7} + 1 \]

5. **Using the Roots of Unity**:
The expression \( z^2 + z + 1 \) can also be interpreted as the sum of the roots of the polynomial \( x^3 - 1 = 0 \) (excluding \( x = 1 \)), which gives us:
\[ z^2 + z + 1 = 0 \quad \Rightarrow \quad z^2 + z + 1 = 0 \]
Therefore, the value of \( z^2 + z + 1 \) is:
\[ z^2 + z + 1 = 0 \]

6. **Final Result**:
Thus, we conclude:
\[ z^{86} + z^{175} + z^{289} = 0 \]

### Final Answer:
\[ \boxed{0} \]