Suppose a,b,c and x real numbers. Let
`Delta=|(1+a,1+ax,1+ax^(2)),(1+b,1+bx,1+bx^(2)),(1+c,1+cx,1+cx^(2))|`
Then `Delta` is independent of

Options

A. a,b,c
B. x
C. a,b,c,x
D. None of these

Answer

To solve the problem, we need to evaluate the determinant given by:

\[ \Delta = \begin{vmatrix} 1 + a & 1 + ax & 1 + ax^2 \\ 1 + b & 1 + bx & 1 + bx^2 \\ 1 + c & 1 + cx & 1 + cx^2 \end{vmatrix} \]

### Step 1: Split the first column
We can split the first column into two parts:

\[ \Delta = \begin{vmatrix} 1 & 1 + ax & 1 + ax^2 \\ 1 & 1 + bx & 1 + bx^2 \\ 1 & 1 + cx & 1 + cx^2 \end{vmatrix} + \begin{vmatrix} a & 1 + ax & 1 + ax^2 \\ b & 1 + bx & 1 + bx^2 \\ c & 1 + cx & 1 + cx^2 \end{vmatrix} \]

### Step 2: Apply column operations
Now, we can perform column operations to simplify the determinant. We will replace the second and third columns by subtracting the first column from them:

\[ \Delta = \begin{vmatrix} 1 & ax & ax^2 \\ 1 & bx & bx^2 \\ 1 & cx & cx^2 \end{vmatrix} + \begin{vmatrix} a & 1 + ax & 1 + ax^2 \\ b & 1 + bx & 1 + bx^2 \\ c & 1 + cx & 1 + cx^2 \end{vmatrix} \]

### Step 3: Further simplify
Now, we can replace the second column with the second column minus \(x\) times the first column and the third column with the third column minus \(x^2\) times the first column:

\[ \Delta = \begin{vmatrix} 1 & ax & ax^2 \\ 1 & bx & bx^2 \\ 1 & cx & cx^2 \end{vmatrix} \]

This results in:

\[ \Delta = \begin{vmatrix} 1 & ax & ax^2 \\ 1 & bx & bx^2 \\ 1 & cx & cx^2 \end{vmatrix} \]

### Step 4: Identify linear dependence
Notice that the first column is the same for all rows. Therefore, the rows are linearly dependent, which implies that the determinant is zero:

\[ \Delta = 0 \]

### Conclusion
Thus, we conclude that the value of \(\Delta\) is independent of \(a\), \(b\), \(c\), and \(x\). The determinant is always zero.