`|(x^(2),y^(2)+z^(2),yz),(y^(2),z^(2)+x^(2),zx),(z^(2),x^(2)+y^(2),xy)|` is divisible by

Options

A. `x^(2)+y^(2)+z^(2)`
B. `x-y`
C. `x-y-z`
D. `x+y+z`

Answer

To solve the determinant

\[ D = \begin{vmatrix} x^2 & y^2 + z^2 & yz \\ y^2 & z^2 + x^2 & zx \\ z^2 & x^2 + y^2 & xy \end{vmatrix} \]

we will perform some operations on the columns to simplify it.

### Step 1: Add Column 1 and Column 2
We start by adding Column 1 to Column 2:

\[ C_2 \rightarrow C_2 + C_1 \]

This gives us:

\[ D = \begin{vmatrix} x^2 & x^2 + y^2 + z^2 & yz \\ y^2 & y^2 + z^2 + x^2 & zx \\ z^2 & z^2 + x^2 + y^2 & xy \end{vmatrix} \]

### Step 2: Factor out the common term from Column 1
Now, we can factor out \(x^2 + y^2 + z^2\) from Column 1:

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

### Step 3: Subtract Rows
Next, we will perform row operations. Subtract Row 2 from Row 1 and Row 3 from Row 2:

\[ R_1 \rightarrow R_1 - R_2 \]
\[ R_2 \rightarrow R_2 - R_3 \]

This results in:

\[ D = (x^2 + y^2 + z^2) \begin{vmatrix} 0 & 0 & yz - zx \\ 0 & 0 & zx - xy \\ z^2 & x^2 + y^2 & xy \end{vmatrix} \]

### Step 4: Expand the Determinant
Since the first two rows are now zeros, we can expand the determinant along the first two rows:

\[ D = (x^2 + y^2 + z^2) \cdot 0 = 0 \]

### Step 5: Identify Divisibility
Now, since we have factored out \(x^2 + y^2 + z^2\), we can conclude that the original determinant is divisible by \(x^2 + y^2 + z^2\).

### Step 6: Check Other Factors
Next, we check for the other factors by performing further row operations and simplifying the determinant. After performing the necessary operations, we find that the determinant is also divisible by \(x - y\), \(y - z\), and \(z - x\).

### Conclusion
Thus, the determinant is divisible by:

- \(x^2 + y^2 + z^2\)
- \(x - y\)
- \(y - z\)
- \(z - x\)

### Final Answer
The determinant is divisible by \(x^2 + y^2 + z^2\), \(x - y\), \(y - z\), and \(z - x\).

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