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 `+- 1`
B. ` 3m +-1`
C. `+-3m `
D. none of these

Answer

To solve the problem step by step, we will analyze the given information and derive the required expression for \( x^3 + y^3 + z^3 \).

### Step 1: Understand the Matrix and Its Properties
We are given the matrix \( P \):
\[ P = \begin{pmatrix} x & y & z \\ z & x & y \\ y & z & x \end{pmatrix} \]
Since \( P \) is an orthogonal matrix, we know that \( P^T P = I \) (the identity matrix).

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

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

Calculating the elements of the resulting matrix:
- The (1,1) entry: \( x^2 + y^2 + z^2 \)
- The (1,2) entry: \( xy + xz + yz \)
- The (1,3) entry: \( xz + yx + zy \)
- The (2,1) entry: \( zx + xy + yz \)
- The (2,2) entry: \( z^2 + x^2 + y^2 \)
- The (2,3) entry: \( zy + zx + xy \)
- The (3,1) entry: \( yx + zy + xz \)
- The (3,2) entry: \( y^2 + z^2 + x^2 \)
- The (3,3) entry: \( z^2 + y^2 + x^2 \)

This gives us:
\[ P P^T = \begin{pmatrix} x^2 + y^2 + z^2 & xy + xz + yz & xy + xz + yz \\ xy + xz + yz & x^2 + y^2 + z^2 & xy + xz + yz \\ xy + xz + yz & xy + xz + yz & x^2 + y^2 + z^2 \end{pmatrix} \]

### Step 4: Set Equal to Identity Matrix
Since \( P P^T = I \), we have:
1. \( x^2 + y^2 + z^2 = 1 \)
2. \( xy + xz + yz = 0 \)

### Step 5: Use the Identity for Cubes
We know the identity:
\[ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - xz - yz) \]
Substituting the known values:
- \( x^2 + y^2 + z^2 = 1 \)
- \( xy + xz + yz = 0 \)

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

### Step 6: Find \( x + y + z \)
From the earlier steps, since \( x^2 + y^2 + z^2 = 1 \) and \( xy + xz + yz = 0 \), we can conclude:
\[ (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + xz + yz) = 1 + 2(0) = 1 \]
Thus, \( x + y + z = \pm 1 \).

### Step 7: Substitute Back
Now substituting back:
- If \( x + y + z = 1 \):
\[ x^3 + y^3 + z^3 = 1 + 3xyz \]
- If \( x + y + z = -1 \):
\[ x^3 + y^3 + z^3 = -1 + 3xyz \]

### Conclusion
Since \( xyz = m \):
\[ x^3 + y^3 + z^3 = 3m \pm 1 \]

Thus, the final answer is:
\[ \boxed{3m \pm 1} \]