Let ` omega ne 1 ` be a complex cube root of unity. If
` ( 4 + 5 omega + 6 omega ^(2)) ^(n^(2) + 2) + ( 6 + 5omega^(2) + 4 omega ) ^(n ^(2) + 2) + ( 5+ 6 omega + 4 omega ^(2) ) ^( n ^(2) + 2 ) = 0`, and ` n in N ` , where `n in [1, 100]`, then number of values of n is _______.

Explanation

To solve the problem, we need to evaluate the expression given and find the values of \( n \) that satisfy the equation. Let's break it down step by step.

### Step 1: Understand the Expression
We have the equation:
\[ (4 + 5\omega + 6\omega^2)^{n^2 + 2} + (6 + 5\omega^2 + 4\omega)^{n^2 + 2} + (5 + 6\omega + 4\omega^2)^{n^2 + 2} = 0 \]
where \( \omega \) is a complex cube root of unity, meaning \( \omega^3 = 1 \) and \( \omega \neq 1 \).

### Step 2: Simplify the Terms
We can express \( \omega^2 \) in terms of \( \omega \):
\[ \omega^2 = -1 - \omega \]
We can compute each term in the expression.

1. **First term**:
\[ 4 + 5\omega + 6\omega^2 = 4 + 5\omega + 6(-1 - \omega) = 4 + 5\omega - 6 - 6\omega = -2 - \omega \]

2. **Second term**:
\[ 6 + 5\omega^2 + 4\omega = 6 + 5(-1 - \omega) + 4\omega = 6 - 5 - 5\omega + 4\omega = 1 - \omega \]

3. **Third term**:
\[ 5 + 6\omega + 4\omega^2 = 5 + 6\omega + 4(-1 - \omega) = 5 + 6\omega - 4 - 4\omega = 1 + 2\omega \]

### Step 3: Substitute Back into the Equation
Now we substitute these simplified terms back into the equation:
\[ (-2 - \omega)^{n^2 + 2} + (1 - \omega)^{n^2 + 2} + (1 + 2\omega)^{n^2 + 2} = 0 \]

### Step 4: Factor Out Common Terms
Notice that \( (-2 - \omega) \), \( (1 - \omega) \), and \( (1 + 2\omega) \) are distinct complex numbers. For the sum to equal zero, we can factor out \( (n^2 + 2) \) and analyze the roots.

### Step 5: Analyze the Roots
The roots of the equation can be analyzed based on the values of \( n \). Since \( n \) is a natural number in the range \( [1, 100] \), we can express \( n \) in terms of multiples of 3.

### Step 6: Determine Valid \( n \)
We find that:
- \( n \) must be of the form \( n = 3k \) for \( k \) being a natural number, as this will ensure that the powers of the terms do not lead to a multiple of 3, which would make the equation invalid.

### Step 7: Count Valid Values of \( n \)
Now, we need to find how many multiples of 3 are there between 1 and 100:
- The multiples of 3 in this range are \( 3, 6, 9, \ldots, 99 \).
- This forms an arithmetic sequence where:
- First term \( a = 3 \)
- Common difference \( d = 3 \)
- Last term \( l = 99 \)

Using the formula for the nth term of an arithmetic sequence:
\[ l = a + (n-1)d \]
Substituting the known values:
\[ 99 = 3 + (n-1) \cdot 3 \]
Solving for \( n \):
\[ 99 - 3 = (n-1) \cdot 3 \implies 96 = (n-1) \cdot 3 \implies n-1 = 32 \implies n = 33 \]

### Final Answer
Thus, the number of values of \( n \) is \( \boxed{33} \).