Prove that `x^(3p) + x^(3q+1) + x^(3r+2)` is exactly divisible by `x^2+x+1`, if p,q,r is integer.

Options

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

Answer

To prove that \( x^{3p} + x^{3q+1} + x^{3r+2} \) is exactly divisible by \( x^2 + x + 1 \) for integers \( p, q, r \), we can follow these steps:

### Step 1: Identify the Roots of the Polynomial
The polynomial \( x^2 + x + 1 \) has roots which are the cube roots of unity, denoted as \( \omega \) and \( \omega^2 \), where \( \omega = e^{2\pi i / 3} \). These roots satisfy the equations:
\[ \omega^3 = 1 \quad \text{and} \quad 1 + \omega + \omega^2 = 0 \]

### Step 2: Substitute \( \omega \) into the Expression
We will evaluate the expression \( x^{3p} + x^{3q+1} + x^{3r+2} \) at \( x = \omega \):
\[ \omega^{3p} + \omega^{3q+1} + \omega^{3r+2} \]

### Step 3: Simplify Each Term
Using the property \( \omega^3 = 1 \):
- \( \omega^{3p} = 1 \)
- \( \omega^{3q+1} = \omega \)
- \( \omega^{3r+2} = \omega^2 \)

Thus, we can rewrite the expression:
\[ \omega^{3p} + \omega^{3q+1} + \omega^{3r+2} = 1 + \omega + \omega^2 \]

### Step 4: Use the Property of Roots
From the property of cube roots of unity, we know:
\[ 1 + \omega + \omega^2 = 0 \]
Therefore, substituting this back, we get:
\[ \omega^{3p} + \omega^{3q+1} + \omega^{3r+2} = 0 \]

### Step 5: Repeat for \( \omega^2 \)
Now, we will evaluate the expression at \( x = \omega^2 \):
\[ (\omega^2)^{3p} + (\omega^2)^{3q+1} + (\omega^2)^{3r+2} \]
Using \( (\omega^2)^3 = 1 \):
- \( (\omega^2)^{3p} = 1 \)
- \( (\omega^2)^{3q+1} = \omega^2 \)
- \( (\omega^2)^{3r+2} = \omega \)

Thus, we can rewrite the expression:
\[ (\omega^2)^{3p} + (\omega^2)^{3q+1} + (\omega^2)^{3r+2} = 1 + \omega^2 + \omega \]
Again, using the property of cube roots of unity:
\[ 1 + \omega^2 + \omega = 0 \]

### Conclusion
Since both \( \omega \) and \( \omega^2 \) are roots of the polynomial \( x^2 + x + 1 \), we conclude that \( x^{3p} + x^{3q+1} + x^{3r+2} \) is exactly divisible by \( x^2 + x + 1 \).