If `a^(2) + b^(2) + c^(2) = -2 and f(x) = |(1 + a^(2)x,(1 + b^(2))x,(1 + c^(2)) x),((1 + a^(2))x,1 + b^(2)x,(1 + c^(2))x),((1 + a^(2)) x,(1 + b^(2))x,1 + c^(2)x)|`, then f(x) is a polynomial of degree

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem step by step, we need to evaluate the determinant given in the function \( f(x) \) and determine its degree.

### Step 1: Write the determinant
The function is given as:
\[ f(x) = \begin{vmatrix} (1 + a^2)x & (1 + b^2)x & (1 + c^2)x \\ (1 + a^2)x & 1 + b^2x & (1 + c^2)x \\ (1 + a^2)x & (1 + b^2)x & 1 + c^2x \end{vmatrix} \]

### Step 2: Simplify the determinant
We can simplify the determinant by performing column operations. We will add all three columns together to the first column:
\[ C_1 \rightarrow C_1 + C_2 + C_3 \]

This gives us:
\[ f(x) = \begin{vmatrix} 3 & 1 + (a^2 + b^2 + c^2)x & (1 + c^2)x \\ 3 & 1 + (a^2 + b^2 + c^2)x & (1 + c^2)x \\ 3 & (1 + b^2)x & 1 + c^2x \end{vmatrix} \]

### Step 3: Substitute the given condition
We know from the problem that \( a^2 + b^2 + c^2 = -2 \). Substituting this into our determinant:
\[ f(x) = \begin{vmatrix} 3 & 1 - 2x & (1 + c^2)x \\ 3 & 1 - 2x & (1 + c^2)x \\ 3 & (1 + b^2)x & 1 + c^2x \end{vmatrix} \]

### Step 4: Further simplify the determinant
Now, we can notice that the first two rows are identical. Therefore, the determinant simplifies to zero:
\[ f(x) = 0 \]

### Step 5: Determine the polynomial degree
Since \( f(x) = 0 \), it is a constant polynomial (specifically, the zero polynomial). The degree of the zero polynomial is defined as:
\[ \text{Degree of } f(x) = -\infty \]

### Final Answer
Thus, the degree of the polynomial \( f(x) \) is:
\[ \text{Degree of } f(x) = -\infty \]