Value of `|{:(1+x_(1),,1+x_(1)x,,1+x_(1)x^(2)),(1+x_(2),,1+x_(2)x,,1+x_(2)x^(2)),(1+x_(3),,1+x_(3)x,,1+x_(3)x^(2)):}|` depends upon

Options

A. `x` only
B. `x_(1)`only
C. `x_(2)`only
D. none of these

Explanation

To solve the determinant

\[ D = \begin{vmatrix} 1 + x_1 & 1 + x_1 x & 1 + x_1 x^2 \\ 1 + x_2 & 1 + x_2 x & 1 + x_2 x^2 \\ 1 + x_3 & 1 + x_3 x & 1 + x_3 x^2 \end{vmatrix} \]

we can analyze it step by step.

### Step 1: Rewrite the Determinant
We can express the determinant in a more manageable form. Notice that each row has a common structure. We can factor out common terms.

### Step 2: Factor Out Common Elements
Let's factor out \(1\) from each row:

\[ D = \begin{vmatrix} 1 + x_1 & 1 + x_1 x & 1 + x_1 x^2 \\ 1 + x_2 & 1 + x_2 x & 1 + x_2 x^2 \\ 1 + x_3 & 1 + x_3 x & 1 + x_3 x^2 \end{vmatrix} \]

This determinant can be simplified by observing that if we subtract the first column from the second and the third columns, we can simplify our calculations.

### Step 3: Perform Column Operations
Subtract the first column from the second and third columns:

\[ D = \begin{vmatrix} 1 + x_1 & x_1 x & x_1 x^2 \\ 1 + x_2 & x_2 x & x_2 x^2 \\ 1 + x_3 & x_3 x & x_3 x^2 \end{vmatrix} \]

### Step 4: Factor Out \(x_1\), \(x_2\), and \(x_3\)
Now, we can factor out \(x_1\), \(x_2\), and \(x_3\) from the second and third columns:

\[ D = \begin{vmatrix} 1 + x_1 & x_1 & x_1 \\ 1 + x_2 & x_2 & x_2 \\ 1 + x_3 & x_3 & x_3 \end{vmatrix} \]

### Step 5: Analyze the Determinant
Now, we can see that if we have two identical columns (the second and third columns are identical), the determinant will be zero:

\[ D = 0 \]

### Conclusion
The value of the determinant \(D\) is zero regardless of the values of \(x_1\), \(x_2\), and \(x_3\). Therefore, the value of the determinant does not depend on \(x_1\), \(x_2\), or \(x_3\).

### Final Answer
The value of the determinant depends on **none of the variables**.

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