`"Let" P=[{:(,"cos"(pi)/(9),"sin"(pi)/(9)),(,-"sin"(pi)/(9),"cos"(pi)/(9)):}]` and `alpha, beta, gamma` be non-zero real number such that `alpha p^(6)+betap^(3)+gammal` is the zero matrix. Then find the value of `(alpha^(2)+beta^(2)+gamma^(2))^(alpha-betaxxbeta-gammaxxgamma-alpha)`

Answer

To solve the problem, we need to analyze the given matrix \( P \) and the equation involving \( \alpha, \beta, \gamma \). Let's break it down step by step.

### Step 1: Define the Matrix \( P \)
The matrix \( P \) is defined as:
\[ P = \begin{pmatrix} \cos\left(\frac{\pi}{9}\right) & \sin\left(\frac{\pi}{9}\right) \\ -\sin\left(\frac{\pi}{9}\right) & \cos\left(\frac{\pi}{9}\right) \end{pmatrix} \]
This is a rotation matrix that represents a rotation by an angle of \( \frac{\pi}{9} \).

### Step 2: Calculate \( P^2 \)
To find \( P^2 \), we multiply \( P \) by itself:
\[ P^2 = P \cdot P = \begin{pmatrix} \cos\left(\frac{\pi}{9}\right) & \sin\left(\frac{\pi}{9}\right) \\ -\sin\left(\frac{\pi}{9}\right) & \cos\left(\frac{\pi}{9}\right) \end{pmatrix} \cdot \begin{pmatrix} \cos\left(\frac{\pi}{9}\right) & \sin\left(\frac{\pi}{9}\right) \\ -\sin\left(\frac{\pi}{9}\right) & \cos\left(\frac{\pi}{9}\right) \end{pmatrix} \]
Using the multiplication of matrices, we find:
\[ P^2 = \begin{pmatrix} \cos\left(\frac{2\pi}{9}\right) & \sin\left(\frac{2\pi}{9}\right) \\ -\sin\left(\frac{2\pi}{9}\right) & \cos\left(\frac{2\pi}{9}\right) \end{pmatrix} \]

### Step 3: Calculate \( P^3 \)
Continuing this process, we can find \( P^3 \):
\[ P^3 = P^2 \cdot P = \begin{pmatrix} \cos\left(\frac{2\pi}{9}\right) & \sin\left(\frac{2\pi}{9}\right) \\ -\sin\left(\frac{2\pi}{9}\right) & \cos\left(\frac{2\pi}{9}\right) \end{pmatrix} \cdot P \]
This results in:
\[ P^3 = \begin{pmatrix} \cos\left(\frac{3\pi}{9}\right) & \sin\left(\frac{3\pi}{9}\right) \\ -\sin\left(\frac{3\pi}{9}\right) & \cos\left(\frac{3\pi}{9}\right) \end{pmatrix} \]

### Step 4: Calculate \( P^6 \)
Following the same logic:
\[ P^6 = P^3 \cdot P^3 = \begin{pmatrix} \cos\left(\frac{6\pi}{9}\right) & \sin\left(\frac{6\pi}{9}\right) \\ -\sin\left(\frac{6\pi}{9}\right) & \cos\left(\frac{6\pi}{9}\right) \end{pmatrix} \]

### Step 5: Set Up the Equation
Given the equation:
\[ \alpha P^6 + \beta P^3 + \gamma I = 0 \]
Substituting the values we calculated:
\[ \alpha \begin{pmatrix} \cos\left(\frac{6\pi}{9}\right) & \sin\left(\frac{6\pi}{9}\right) \\ -\sin\left(\frac{6\pi}{9}\right) & \cos\left(\frac{6\pi}{9}\right) \end{pmatrix} + \beta \begin{pmatrix} \cos\left(\frac{3\pi}{9}\right) & \sin\left(\frac{3\pi}{9}\right) \\ -\sin\left(\frac{3\pi}{9}\right) & \cos\left(\frac{3\pi}{9}\right) \end{pmatrix} + \gamma \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = 0 \]

### Step 6: Solve for \( \alpha, \beta, \gamma \)
From the equation, we can derive the following relationships:
1. The coefficients of the matrix must equal zero.
2. This gives us a system of equations to solve.

### Step 7: Find \( \alpha, \beta, \gamma \)
From the relationships derived, we find:
- \( \alpha + \beta = 0 \) implies \( \beta = -\alpha \)
- \( \alpha - \beta = 2\gamma \) implies \( 2\alpha = 2\gamma \) or \( \alpha = \gamma \)

### Step 8: Substitute Values
Substituting \( \beta = -\alpha \) and \( \gamma = \alpha \) into \( \alpha^2 + \beta^2 + \gamma^2 \):
\[ \alpha^2 + (-\alpha)^2 + \alpha^2 = 3\alpha^2 \]

### Step 9: Calculate the Exponent
The exponent simplifies as follows:
\[ \alpha - \beta = 2\alpha, \quad \beta - \gamma = -2\alpha, \quad \gamma - \alpha = 0 \]
Thus, the entire exponent becomes:
\[ (2\alpha)(-2\alpha)(0) = 0 \]

### Step 10: Final Result
Since any non-zero number raised to the power of 0 is 1:
\[ (3\alpha^2)^0 = 1 \]

### Final Answer
The value is:
\[ \boxed{1} \]