If `Delta=|(1+alpha,1+alphax,1+ax^2),(1+beta,1+betax,1+betax^2),(1+gamma,1+gammax,1+gammax^2)|` then `Delta =`

Options

A. 0
B. `(alpha - beta)(beta-gamma)(gamma-alpha)`
C. `alpha beta gamma`
D. none

Answer

To solve the determinant given by

\[ \Delta = \begin{vmatrix} 1 + \alpha & 1 + \alpha x & 1 + \alpha x^2 \\ 1 + \beta & 1 + \beta x & 1 + \beta x^2 \\ 1 + \gamma & 1 + \gamma x & 1 + \gamma x^2 \end{vmatrix}, \]

we will apply row operations to simplify the determinant.

### Step 1: Apply Row Operations
We will perform the following row operations:
- \( R_1 \leftarrow R_1 - R_3 \)
- \( R_2 \leftarrow R_2 - R_3 \)

This means we will subtract the third row from the first and second rows.

### Step 2: Calculate New Rows
After performing the row operations, we get:

1. For \( R_1 \):
\[ R_1 = (1 + \alpha - (1 + \gamma), 1 + \alpha x - (1 + \gamma x), 1 + \alpha x^2 - (1 + \gamma x^2)) \]
Simplifying this gives:
\[ R_1 = (\alpha - \gamma, \alpha - \gamma x, \alpha - \gamma x^2) \]

2. For \( R_2 \):
\[ R_2 = (1 + \beta - (1 + \gamma), 1 + \beta x - (1 + \gamma x), 1 + \beta x^2 - (1 + \gamma x^2)) \]
Simplifying this gives:
\[ R_2 = (\beta - \gamma, \beta - \gamma x, \beta - \gamma x^2) \]

The third row remains unchanged:
\[ R_3 = (1 + \gamma, 1 + \gamma x, 1 + \gamma x^2) \]

### Step 3: Write the New Determinant
Now the determinant looks like:
\[ \Delta = \begin{vmatrix} \alpha - \gamma & \alpha - \gamma x & \alpha - \gamma x^2 \\ \beta - \gamma & \beta - \gamma x & \beta - \gamma x^2 \\ 1 + \gamma & 1 + \gamma x & 1 + \gamma x^2 \end{vmatrix} \]

### Step 4: Factor Out Common Terms
Notice that:
- In the first row, we can factor out \( \alpha - \gamma \).
- In the second row, we can factor out \( \beta - \gamma \).

Thus, we can express the determinant as:
\[ \Delta = (\alpha - \gamma)(\beta - \gamma) \begin{vmatrix} 1 & 1 & 1 \\ 1 & x & x^2 \\ 1 + \gamma & 1 + \gamma x & 1 + \gamma x^2 \end{vmatrix} \]

### Step 5: Recognize Row Dependency
Now, observe that the first two rows of the new determinant are linearly dependent. The first row is a constant row, while the second row is a polynomial row. Therefore, the determinant evaluates to zero:
\[ \Delta = 0 \]

### Final Answer
Thus, the value of the determinant \( \Delta \) is:
\[ \Delta = 0 \]