The repeated factor of the determinant
`|(y +z,x,y),(z +x,z,x),(x +y,y,z)|`, is

Options

A. `x-z`
B. `x -y`
C. `y -z`
D. none of these

Answer

To find the repeated factor of the determinant

\[ D = \begin{vmatrix} y + z & x & y \\ z + x & z & x \\ x + y & y & z \end{vmatrix} \]

we will perform some row operations and simplify the determinant step by step.

### Step 1: Apply Row Operation
We will add all three rows together and replace the first row with this sum.

\[ R_1 \rightarrow R_1 + R_2 + R_3 \]

Calculating the new first row:

- First element: \((y + z) + (z + x) + (x + y) = 2x + 2y + 2z = 2(x + y + z)\)
- Second element: \(x + z + y = x + y + z\)
- Third element: \(y + x + z = x + y + z\)

So, the determinant becomes:

\[ D = \begin{vmatrix} 2(x + y + z) & x + y + z & x + y + z \\ z + x & z & x \\ x + y & y & z \end{vmatrix} \]

### Step 2: Factor Out Common Terms
Notice that the first row has a common factor of \(x + y + z\):

\[ D = (x + y + z) \begin{vmatrix} 2 & 1 & 1 \\ z + x & z & x \\ x + y & y & z \end{vmatrix} \]

### Step 3: Expand the Determinant
Now we will expand the determinant:

\[ D = (x + y + z) \left[ 2 \begin{vmatrix} z & x \\ y & z \end{vmatrix} - 1 \begin{vmatrix} z + x & x \\ x + y & z \end{vmatrix} + 1 \begin{vmatrix} z + x & z \\ x + y & y \end{vmatrix} \right] \]

Calculating the smaller determinants:

1. \(\begin{vmatrix} z & x \\ y & z \end{vmatrix} = z^2 - xy\)

2. \(\begin{vmatrix} z + x & x \\ x + y & z \end{vmatrix} = (z + x)z - x(x + y) = z^2 + xz - x^2 - xy\)

3. \(\begin{vmatrix} z + x & z \\ x + y & y \end{vmatrix} = (z + x)y - z(x + y) = zy + xy - zx - zy = xy - zx\)

### Step 4: Substitute Back
Substituting these back into the determinant:

\[ D = (x + y + z) \left[ 2(z^2 - xy) - (z^2 + xz - x^2 - xy) + (xy - zx) \right] \]

### Step 5: Simplify
Now we simplify:

\[ = (x + y + z) \left[ 2z^2 - 2xy - z^2 - xz + x^2 + xy + xy - zx \right] \]

Combining like terms:

\[ = (x + y + z) \left[ z^2 - xz + x^2 - 2xy \right] \]

### Step 6: Factor Further
Notice that \(z^2 - xz + x^2 - 2xy\) can be rewritten as:

\[ = (x - z)^2 \]

### Final Result
Thus, we have:

\[ D = (x + y + z)(x - z)^2 \]

The repeated factor of the determinant is:

\[ \boxed{(x - z)} \]