` " if " Delta = |{:(-x,,a,,b),(b,,-x,,a),(a,,b,,-x):}|" then a factor of " Delta " is "`

Options

A. `a+b+x`
B. `x^(2)-(a-b)x+a^(2)+b^(2)+ab`
C. `x^(2)+(a+b)x+a^(2)+b^(2)-ab`
D. `a+b-x`

Explanation

To solve the given problem, we need to find a factor of the determinant \( \Delta = \begin{vmatrix} -x & a & b \\ b & -x & a \\ a & b & -x \end{vmatrix} \).

### Step-by-Step Solution:

1. **Write the Determinant**:
\[ \Delta = \begin{vmatrix} -x & a & b \\ b & -x & a \\ a & b & -x \end{vmatrix} \]

2. **Apply Column Operations**:
We can simplify the determinant by applying the operation \( C_1 \to C_1 + C_2 + C_3 \):
\[ C_1 = -x + a + b, \quad C_2 = a, \quad C_3 = b \]
This gives us:
\[ \Delta = \begin{vmatrix} a + b - x & a & b \\ b & -x & a \\ a & b & -x \end{vmatrix} \]

3. **Row Operations**:
Next, we perform row operations to simplify further. We can apply \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
\[ R_2 = \begin{pmatrix} b - (a + b - x) & -x - a & a - b \end{pmatrix} = \begin{pmatrix} -a + x & -x - a & a - b \end{pmatrix} \]
\[ R_3 = \begin{pmatrix} a - (a + b - x) & b - a & -x - b \end{pmatrix} = \begin{pmatrix} -b + x & b - a & -x - b \end{pmatrix} \]
So, we have:
\[ \Delta = \begin{vmatrix} a + b - x & a & b \\ -a + x & -x - a & a - b \\ -b + x & b - a & -x - b \end{vmatrix} \]

4. **Factor Out Common Terms**:
We can factor out \( a + b - x \) from the first row:
\[ \Delta = (a + b - x) \begin{vmatrix} 1 & a & b \\ -a + x & -x - a & a - b \\ -b + x & b - a & -x - b \end{vmatrix} \]

5. **Expand the Determinant**:
Now, we can expand the determinant along the first row. This will yield a polynomial in terms of \( x \).

6. **Final Factorization**:
After performing the expansion and simplification, we will find that:
\[ \Delta = (a + b - x)(x^2 + (a + b)x + (a^2 + b^2 - ab)) \]
Therefore, one of the factors of \( \Delta \) is \( a + b - x \).

### Conclusion:
The factor of \( \Delta \) is \( a + b - x \).