The factors of `|(x,a,b),(a,x,b),(a,b,x)|`, are

Options

A. `x - a, x -b, and x + a + b`
B. `x + a, x + b and x + a + b`
C. `x + a, x + b and x - a -b`
D. `x - a, x - b and x - a -b`

Answer

To find the factors of the determinant \( |(x,a,b),(a,x,b),(a,b,x)| \), we can follow these steps:

### Step 1: Write the Determinant
We start with the determinant:
\[ D = \begin{vmatrix} x & a & b \\ a & x & b \\ a & b & x \end{vmatrix} \]

### Step 2: Apply Column Operations
We can simplify the determinant by performing column operations. Let's add all three columns together into the first column:
\[ C_1 = C_1 + C_2 + C_3 \]
This gives us:
\[ D = \begin{vmatrix} x + a + b & a & b \\ a + x + b & x & b \\ a + b + x & b & x \end{vmatrix} \]
Now, we can denote \( S = x + a + b \), so the matrix becomes:
\[ D = \begin{vmatrix} S & a & b \\ S & x & b \\ S & b & x \end{vmatrix} \]

### Step 3: Subtract Rows
Next, we can subtract the first row from the second and third rows:
\[ R_2 = R_2 - R_1 \quad \text{and} \quad R_3 = R_3 - R_1 \]
This gives us:
\[ D = \begin{vmatrix} S & a & b \\ 0 & x - a & 0 \\ 0 & b - a & x - b \end{vmatrix} \]

### Step 4: Expand the Determinant
Now we can expand the determinant along the first column:
\[ D = S \cdot \begin{vmatrix} x - a & 0 \\ b - a & x - b \end{vmatrix} \]
Calculating the 2x2 determinant:
\[ D = S \cdot ((x - a)(x - b) - 0) = S \cdot (x - a)(x - b) \]

### Step 5: Substitute Back for S
Substituting back \( S = x + a + b \):
\[ D = (x + a + b)(x - a)(x - b) \]

### Conclusion
Thus, the factors of the determinant \( |(x,a,b),(a,x,b),(a,b,x)| \) are:
\[ (x + a + b)(x - a)(x - b) \]