The expression ` x^(3p) + x^(3q - 1) + x^(3r - 2)` , where ` p, q, r in N` is divisible by

Options

A. `x^(2) - x + 1 `
B. ` x^(2) + x + 1 `
C. ` x^(2) + x - 1 `
D. `x^(2) - x - 1 `

Answer

To solve the expression \( x^{3p} + x^{3q - 1} + x^{3r - 2} \) where \( p, q, r \in \mathbb{N} \), we will analyze the powers of \( x \) using the properties of complex cube roots of unity.

### Step 1: Understand the properties of \( x \)
Let \( x \) be a complex cube root of unity. The cube roots of unity are given by:
\[ x = e^{2\pi i / 3}, \quad x^2 = e^{4\pi i / 3}, \quad x^3 = 1 \]
Thus, for any integer \( n \), we have:
\[ x^n = x^{n \mod 3} \]

### Step 2: Simplify each term in the expression
1. For \( x^{3p} \):
\[ x^{3p} = (x^3)^p = 1^p = 1 \]

2. For \( x^{3q - 1} \):
\[ x^{3q - 1} = \frac{x^{3q}}{x} = \frac{1}{x} \]

3. For \( x^{3r - 2} \):
\[ x^{3r - 2} = \frac{x^{3r}}{x^2} = \frac{1}{x^2} \]

### Step 3: Combine the simplified terms
Now, substituting these results back into the original expression:
\[ x^{3p} + x^{3q - 1} + x^{3r - 2} = 1 + \frac{1}{x} + \frac{1}{x^2} \]

### Step 4: Find a common denominator
The common denominator for the terms \( 1, \frac{1}{x}, \frac{1}{x^2} \) is \( x^2 \):
\[ 1 = \frac{x^2}{x^2}, \quad \frac{1}{x} = \frac{x}{x^2}, \quad \frac{1}{x^2} = \frac{1}{x^2} \]
Thus, we can rewrite the expression as:
\[ \frac{x^2 + x + 1}{x^2} \]

### Step 5: Analyze the numerator
The numerator \( x^2 + x + 1 \) is a polynomial that can be factored or analyzed further. Since \( x \) is a cube root of unity, we know that:
\[ x^3 - 1 = (x - 1)(x^2 + x + 1) \]
This implies that \( x^2 + x + 1 = 0 \) has roots at the non-real cube roots of unity.

### Conclusion
The expression \( x^{3p} + x^{3q - 1} + x^{3r - 2} \) simplifies to:
\[ \frac{x^2 + x + 1}{x^2} \]
Since \( x^2 + x + 1 = 0 \) for \( x \) being a non-real cube root of unity, we conclude that the original expression is divisible by \( x^2 + x + 1 \).

### Final Answer
Thus, the expression \( x^{3p} + x^{3q - 1} + x^{3r - 2} \) is divisible by \( x^2 + x + 1 \).

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