Let `A=[(x,y,z),(y,z,x),(z,x,y)]`, where x, y and z are real numbers such that `x + y + z > 0 and xyz = 2`. If `A^(2) = I_(3)`, then the value of `x^(3)+y^(3)+z^(3)` is__________.

Answer

To solve the problem, we start with the matrix \( A = \begin{pmatrix} x & y & z \\ y & z & x \\ z & x & y \end{pmatrix} \) and the conditions \( x + y + z > 0 \) and \( xyz = 2 \). We also know that \( A^2 = I_3 \), where \( I_3 \) is the identity matrix of order 3.

### Step 1: Understanding the condition \( A^2 = I_3 \)

Since \( A^2 = I_3 \), we can infer that \( A \) is an orthogonal matrix. This means that \( A^T A = I_3 \), where \( A^T \) is the transpose of \( A \).

### Step 2: Calculate the transpose of \( A \)

The transpose of matrix \( A \) is:
\[ A^T = \begin{pmatrix} x & y & z \\ y & z & x \\ z & x & y \end{pmatrix} \]
Since \( A = A^T \), it confirms that \( A \) is symmetric.

### Step 3: Set up the equation \( A A^T = I_3 \)

We can now compute \( A A^T \):
\[ A A^T = \begin{pmatrix} x & y & z \\ y & z & x \\ z & x & y \end{pmatrix} \begin{pmatrix} x & y & z \\ y & z & x \\ z & x & y \end{pmatrix} \]

Calculating the elements of \( A A^T \):
- The (1,1) entry: \( x^2 + y^2 + z^2 \)
- The (1,2) entry: \( xy + yz + zx \)
- The (1,3) entry: \( xz + yx + zy \) (which is the same as (1,2))
- The (2,1) entry: \( yx + zy + xz \) (which is the same as (1,2))
- The (2,2) entry: \( y^2 + z^2 + x^2 \)
- The (2,3) entry: \( yz + zx + xy \) (which is the same as (1,2))
- The (3,1) entry: \( zx + xy + yz \) (which is the same as (1,2))
- The (3,2) entry: \( zy + x^2 + y^2 \) (which is the same as (2,2))
- The (3,3) entry: \( z^2 + x^2 + y^2 \)

Thus, we have:
\[ A A^T = \begin{pmatrix} x^2 + y^2 + z^2 & xy + yz + zx & xy + yz + zx \\ xy + yz + zx & y^2 + z^2 + x^2 & xy + yz + zx \\ xy + yz + zx & xy + yz + zx & z^2 + x^2 + y^2 \end{pmatrix} \]

### Step 4: Set the equations from \( A A^T = I_3 \)

From \( A A^T = I_3 \), we have:
1. \( x^2 + y^2 + z^2 = 1 \)
2. \( xy + yz + zx = 0 \)

### Step 5: Use the identity for cubes

We can use the identity:
\[ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \]

Substituting \( xyz = 2 \), we have:
\[ x^3 + y^3 + z^3 - 6 = (x + y + z)(1 - 0) \]
Thus:
\[ x^3 + y^3 + z^3 = (x + y + z) + 6 \]

### Step 6: Find \( x + y + z \)

From the earlier equations, we know:
\[ x + y + z = 1 \]
Substituting this back, we get:
\[ x^3 + y^3 + z^3 = 1 + 6 = 7 \]

### Final Answer

The value of \( x^3 + y^3 + z^3 \) is \( \boxed{7} \).