`x^(3m) + x^(3n-1) + x^(3r-2)`, where, `m,n,r in N` is divisible by

Options

A. m,n,k are rational
B. m,n,k are integers
C. m,n,k are positive integers
D. none of these

Answer

To determine the expression \( x^{3m} + x^{3n-1} + x^{3r-2} \) where \( m, n, r \in \mathbb{N} \) is divisible by, we can analyze the expression in terms of the cube roots of unity.

### Step-by-Step Solution:

1. **Understanding the Expression**:
We have the expression \( x^{3m} + x^{3n-1} + x^{3r-2} \). Here, \( m, n, r \) are natural numbers.

2. **Using Cube Roots of Unity**:
The cube roots of unity are \( \omega = e^{2\pi i / 3} \) and \( \omega^2 = e^{-2\pi i / 3} \), where \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \).

3. **Substituting \( x = \omega \)**:
Let's substitute \( x = \omega \) into the expression:
\[ \omega^{3m} + \omega^{3n-1} + \omega^{3r-2} \]
Since \( \omega^3 = 1 \), we have:
\[ \omega^{3m} = 1, \quad \omega^{3n-1} = \omega^{-1} = \omega^2, \quad \omega^{3r-2} = \omega^{-2} = \omega \]
Therefore, the expression becomes:
\[ 1 + \omega^2 + \omega \]
Using the property of cube roots of unity:
\[ 1 + \omega + \omega^2 = 0 \]
Thus, substituting gives:
\[ 1 + \omega + \omega^2 = 0 \Rightarrow 0 \]

4. **Substituting \( x = \omega^2 \)**:
Now, substitute \( x = \omega^2 \):
\[ (\omega^2)^{3m} + (\omega^2)^{3n-1} + (\omega^2)^{3r-2} \]
Again, using \( \omega^3 = 1 \):
\[ (\omega^2)^{3m} = 1, \quad (\omega^2)^{3n-1} = \omega, \quad (\omega^2)^{3r-2} = \omega^2 \]
Thus, the expression becomes:
\[ 1 + \omega + \omega^2 \]
Again, we find:
\[ 1 + \omega + \omega^2 = 0 \]

5. **Forming the Polynomial**:
Since both \( \omega \) and \( \omega^2 \) are roots of the expression, we can form the polynomial:
\[ (x - \omega)(x - \omega^2) = x^2 - (\omega + \omega^2)x + \omega \cdot \omega^2 \]
Since \( \omega + \omega^2 = -1 \) and \( \omega \cdot \omega^2 = 1 \):
\[ x^2 + x + 1 \]

6. **Conclusion**:
Therefore, the expression \( x^{3m} + x^{3n-1} + x^{3r-2} \) is divisible by:
\[ x^2 + x + 1 \]