If xyz=m and det `p=|{:(x,y,z),(z,x,y),(y,z,x):}|` , where p is an orthogonal matrix.
The value of `x^3+y^3+z^3` can be

Options

A. `6m pm1`
B. `3m pm1`
C. `pm 3m`
D. none of these

Answer

To solve the problem step by step, we need to analyze the given information and use the properties of determinants and orthogonal matrices.

### Step 1: Understand the Orthogonal Matrix
Given that \( P = \begin{pmatrix} x & y & z \\ z & x & y \\ y & z & x \end{pmatrix} \) is an orthogonal matrix, we know that \( P^T P = I \), where \( P^T \) is the transpose of \( P \) and \( I \) is the identity matrix.

### Step 2: Calculate the Transpose of Matrix P
The transpose of matrix \( P \) is:
\[ P^T = \begin{pmatrix} x & z & y \\ y & x & z \\ z & y & x \end{pmatrix} \]

### Step 3: Multiply \( P \) and \( P^T \)
We need to compute \( P P^T \):
\[ P P^T = \begin{pmatrix} x & y & z \\ z & x & y \\ y & z & x \end{pmatrix} \begin{pmatrix} x & z & y \\ y & x & z \\ z & y & x \end{pmatrix} \]

Calculating the elements of the resulting matrix:
- First row, first column: \( x^2 + y^2 + z^2 \)
- First row, second column: \( xy + xz + yz \)
- First row, third column: \( xz + yz + xy \)

Continuing this for all rows, we get:
\[ P P^T = \begin{pmatrix} x^2 + y^2 + z^2 & xy + xz + yz & xy + xz + yz \\ xy + xz + yz & z^2 + x^2 + y^2 & xy + xz + yz \\ xy + xz + yz & xy + xz + yz & y^2 + z^2 + x^2 \end{pmatrix} \]

### Step 4: Set Equal to Identity Matrix
Since \( P \) is orthogonal, we set \( P P^T = I \):
\[ \begin{pmatrix} x^2 + y^2 + z^2 & xy + xz + yz & xy + xz + yz \\ xy + xz + yz & z^2 + x^2 + y^2 & xy + xz + yz \\ xy + xz + yz & xy + xz + yz & y^2 + z^2 + x^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

From this, we can derive:
1. \( x^2 + y^2 + z^2 = 1 \)
2. \( xy + xz + yz = 0 \)

### Step 5: Use the Identity for \( x^3 + y^3 + z^3 \)
We can use the identity:
\[ x^3 + y^3 + z^3 = (x + y + z)(x^2 + y^2 + z^2 - xy - xz - yz) + 3xyz \]

### Step 6: Substitute Known Values
From our earlier findings:
- \( x^2 + y^2 + z^2 = 1 \)
- \( xy + xz + yz = 0 \)

Thus:
\[ x^3 + y^3 + z^3 = (x + y + z)(1 - 0) + 3xyz = (x + y + z) + 3xyz \]

### Step 7: Find \( x + y + z \)
Using the equation \( (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + xz + yz) \):
\[ (x + y + z)^2 = 1 + 2(0) = 1 \]
Thus, \( x + y + z = \pm 1 \).

### Step 8: Substitute Back
Let \( xyz = m \):
\[ x^3 + y^3 + z^3 = (x + y + z) + 3m \]
So we have:
\[ x^3 + y^3 + z^3 = \pm 1 + 3m \]

### Final Answer
The value of \( x^3 + y^3 + z^3 \) can be \( 3m + 1 \) or \( 3m - 1 \).