If `|(p+x,p,x),(p-x,p,x),(p-x,p,-x)|` = 0 then x is

Options

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

Answer

To solve the determinant equation \( |(p+x, p, x), (p-x, p, x), (p-x, p, -x)| = 0 \), we will proceed step by step.

### Step 1: Write the determinant
We start with the determinant:
\[ D = \begin{vmatrix} p+x & p & x \\ p-x & p & x \\ p-x & p & -x \end{vmatrix} \]

### Step 2: Apply column operations
We can simplify the determinant by performing column operations. Let's replace the first column \( C_1 \) with \( C_1 - C_2 \):
\[ D = \begin{vmatrix} (p+x) - p & p & x \\ (p-x) - p & p & x \\ (p-x) - p & p & -x \end{vmatrix} \]
This simplifies to:
\[ D = \begin{vmatrix} x & p & x \\ -x & p & x \\ -x & p & -x \end{vmatrix} \]

### Step 3: Factor out common terms
Next, we can factor out common terms from the columns:
\[ D = x \begin{vmatrix} 1 & p & 1 \\ -1 & p & 1 \\ -1 & p & -1 \end{vmatrix} \]

### Step 4: Calculate the 3x3 determinant
Now we need to calculate the determinant:
\[ D = x \begin{vmatrix} 1 & p & 1 \\ -1 & p & 1 \\ -1 & p & -1 \end{vmatrix} \]

Using the determinant expansion along the first row:
\[ D = x \left( 1 \cdot \begin{vmatrix} p & 1 \\ p & -1 \end{vmatrix} - p \cdot \begin{vmatrix} -1 & 1 \\ -1 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} -1 & p \\ -1 & p \end{vmatrix} \right) \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} p & 1 \\ p & -1 \end{vmatrix} = p(-1) - p(1) = -p - p = -2p \)
2. \( \begin{vmatrix} -1 & 1 \\ -1 & -1 \end{vmatrix} = (-1)(-1) - (-1)(1) = 1 + 1 = 2 \)
3. \( \begin{vmatrix} -1 & p \\ -1 & p \end{vmatrix} = (-1)(p) - (-1)(p) = -p + p = 0 \)

Substituting these back into the determinant:
\[ D = x \left( 1 \cdot (-2p) - p \cdot 2 + 1 \cdot 0 \right) = x \left( -2p - 2p \right) = x(-4p) \]

### Step 5: Set the determinant equal to zero
We have:
\[ D = -4px = 0 \]
This implies either \( x = 0 \) or \( p = 0 \).

### Conclusion
Since we are asked to find the value of \( x \), the solution is:
\[ \boxed{0} \]