If `z_(1),z_(2),z_(3)` are three complex numbers, such that `|z_(1)|=|z_(2)|=|z_(3)|=1` & `z_(1)^(2)+z_(2)^(2)+z_(3)^(2)=0` then `|z_(1)^(3)+z_(2)^(3)+z_(3)^(3)|` is equal to _______. (not equal to 1)

Explanation

To solve the problem, we start with the given conditions:

1. \( |z_1| = |z_2| = |z_3| = 1 \)
2. \( z_1^2 + z_2^2 + z_3^2 = 0 \)

We need to find \( |z_1^3 + z_2^3 + z_3^3| \).

### Step-by-step Solution:

**Step 1: Express \( z_1, z_2, z_3 \) in terms of a common complex number.**

Since \( |z_1| = |z_2| = |z_3| = 1 \), we can express these complex numbers as:
- \( z_1 = e^{i\theta_1} \)
- \( z_2 = e^{i\theta_2} \)
- \( z_3 = e^{i\theta_3} \)

**Hint:** Use the fact that complex numbers on the unit circle can be expressed in exponential form.

---

**Step 2: Use the condition \( z_1^2 + z_2^2 + z_3^2 = 0 \).**

Substituting the expressions from Step 1:
\[ e^{2i\theta_1} + e^{2i\theta_2} + e^{2i\theta_3} = 0 \]

This implies that the angles \( 2\theta_1, 2\theta_2, 2\theta_3 \) are evenly spaced around the unit circle. Therefore, we can let:
- \( 2\theta_1 = 0 \)
- \( 2\theta_2 = \frac{2\pi}{3} \)
- \( 2\theta_3 = \frac{4\pi}{3} \)

Thus, we have:
- \( z_1 = e^{i \cdot 0} = 1 \)
- \( z_2 = e^{i \cdot \frac{2\pi}{3}} = \omega \)
- \( z_3 = e^{i \cdot \frac{4\pi}{3}} = \omega^2 \)

where \( \omega = e^{i \cdot \frac{2\pi}{3}} \) is a primitive cube root of unity.

**Hint:** Recognize that the sum of the cube roots of unity is zero.

---

**Step 3: Calculate \( z_1^3 + z_2^3 + z_3^3 \).**

Using the property of cube roots of unity:
\[ z_1^3 = 1^3 = 1, \quad z_2^3 = \omega^3 = 1, \quad z_3^3 = (\omega^2)^3 = 1 \]
Thus,
\[ z_1^3 + z_2^3 + z_3^3 = 1 + 1 + 1 = 3 \]

**Hint:** Remember that \( \omega^3 = 1 \) for cube roots of unity.

---

**Step 4: Find the modulus.**

Since \( z_1^3 + z_2^3 + z_3^3 = 3 \), we have:
\[ |z_1^3 + z_2^3 + z_3^3| = |3| = 3 \]

### Final Answer:
The value of \( |z_1^3 + z_2^3 + z_3^3| \) is \( 3 \).

---