The value of the determinant `|{:(m,n,p),(p,m,n),(n,p,m):}|`

Options

A. is a perfect cube
B. is a perfect square
C. has linear factor
D. is zero

Explanation

To find the value of the determinant \( D = \begin{vmatrix} m & n & p \\ p & m & n \\ n & p & m \end{vmatrix} \), we will follow these steps:

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

### Step 2: Perform column operations
We will add all the columns to the first column. This means we will replace the first column with the sum of all three columns:
\[ C_1 \rightarrow C_1 + C_2 + C_3 \]
This gives us:
\[ D = \begin{vmatrix} m+n+p & n & p \\ p+m+n & m & n \\ n+p+m & p & m \end{vmatrix} \]
Now simplifying the first column:
\[ D = \begin{vmatrix} m+n+p & n & p \\ m+n+p & m & n \\ m+n+p & p & m \end{vmatrix} \]

### Step 3: Factor out the common term
Notice that the first column has a common factor of \( m+n+p \). We can factor this out:
\[ D = (m+n+p) \begin{vmatrix} 1 & n & p \\ 1 & m & n \\ 1 & p & m \end{vmatrix} \]

### Step 4: Calculate the remaining determinant
Now we need to calculate the determinant:
\[ \begin{vmatrix} 1 & n & p \\ 1 & m & n \\ 1 & p & m \end{vmatrix} \]
Using the formula for a 3x3 determinant:
\[ = 1 \cdot (m \cdot m - n \cdot p) - n \cdot (1 \cdot m - 1 \cdot p) + p \cdot (1 \cdot n - 1 \cdot p) \]
This simplifies to:
\[ = m^2 - np - n(m - p) + p(n - p) \]
\[ = m^2 - np - nm + np + pn - p^2 \]
\[ = m^2 - nm - p^2 + pn \]

### Step 5: Combine the results
Thus, we have:
\[ D = (m+n+p)(m^2 - nm - p^2 + pn) \]

### Final Result
The value of the determinant is:
\[ D = (m+n+p)(m^2 - nm - p^2 + pn) \]