In the following, each question has one or more than one correct answers. Indicate the correct answer(s). : IF a, b, c are multiples of 3 then `x^a+x^(b+1)+x^(c+2)` is divisible by

Options

A. x+1
B. `x^2+1`
C. `x^2+x+1`
D. none of these

Answer

To solve the problem, we need to determine if the expression \(x^a + x^{b+1} + x^{c+2}\) is divisible by the given options, given that \(a\), \(b\), and \(c\) are multiples of 3.

### Step 1: Understanding the Expression
Given that \(a\), \(b\), and \(c\) are multiples of 3, we can express them as:
- \(a = 3m\)
- \(b = 3n\)
- \(c = 3k\)

Thus, we can rewrite the expression as:
\[ x^{3m} + x^{3n + 1} + x^{3k + 2} \]

### Step 2: Check Divisibility by \(x + 1\)
To check if the expression is divisible by \(x + 1\), we substitute \(x = -1\):
\[ (-1)^{3m} + (-1)^{3n + 1} + (-1)^{3k + 2} \]
Since \(3m\), \(3n\) are even, we have:
\[ 1 + (-1) + 1 = 1 \neq 0 \]
Thus, the expression is **not divisible** by \(x + 1\).

### Step 3: Check Divisibility by \(x^2 + 1\)
Next, we check if it's divisible by \(x^2 + 1\) by substituting \(x = i\) (where \(i\) is the imaginary unit):
\[ i^{3m} + i^{3n + 1} + i^{3k + 2} \]
Calculating each term:
- \(i^{3m} = (i^4)^{m} \cdot i^3 = 1^m \cdot (-i) = -i\)
- \(i^{3n + 1} = (i^4)^{n} \cdot i = 1^n \cdot i = i\)
- \(i^{3k + 2} = (i^4)^{k} \cdot i^2 = 1^k \cdot (-1) = -1\)

Putting it all together:
\[ -i + i - 1 = -1 \neq 0 \]
Thus, the expression is **not divisible** by \(x^2 + 1\).

### Step 4: Check Divisibility by \(x^2 + x + 1\)
Now, we check if it is divisible by \(x^2 + x + 1\). The roots of this polynomial are the non-real cube roots of unity, denoted as \(\omega\) and \(\omega^2\), where:
\[ \omega = e^{2\pi i / 3}, \quad \omega^2 = e^{-2\pi i / 3} \]
We check for \(\omega\):
\[ \omega^{3m} + \omega^{3n + 1} + \omega^{3k + 2} \]
Using the properties of \(\omega\):
\[ \omega^{3m} = 1, \quad \omega^{3n + 1} = \omega, \quad \omega^{3k + 2} = \omega^2 \]
Thus, we have:
\[ 1 + \omega + \omega^2 = 0 \]
This shows that \(\omega\) is a root.

Now, checking for \(\omega^2\):
\[ (\omega^2)^{3m} + (\omega^2)^{3n + 1} + (\omega^2)^{3k + 2} \]
This simplifies to:
\[ 1 + \omega^2 + \omega = 0 \]
Thus, \(\omega^2\) is also a root.

### Conclusion
Since both \(\omega\) and \(\omega^2\) are roots, the expression \(x^a + x^{b+1} + x^{c+2}\) is divisible by \(x^2 + x + 1\).

### Final Answer
The correct answer is:
- **C: \(x^2 + x + 1\)**