Let m be the minimum possible value of `log_(3)(3^(y_(1))+3^(y_(2))+3^(y_(3)))`. Where `y_(1),y_(2),y_(3)` are real number for which `y_(1)+y_(2)+y_(3)=9` Let M be the maxmum possible value of `(log3x_(1)+log_(3) x_(2)+ log_(3) x_(2))` where `x_(1), x_(2), x_(3)` are positive real numbers for whcih `x_(1)+x_(2)+x_(3)=9` then the value of `log_(2)m^(3)+log_(3)(M^(2))` is _____

Answer

To solve the problem, we need to find the minimum value \( m \) of \( \log_3(3^{y_1} + 3^{y_2} + 3^{y_3}) \) given the constraint \( y_1 + y_2 + y_3 = 9 \), and the maximum value \( M \) of \( \log_3(x_1) + \log_3(x_2) + \log_3(x_3) \) given \( x_1 + x_2 + x_3 = 9 \). Finally, we will compute \( \log_2(m^3) + \log_3(M^2) \).

### Step 1: Finding the minimum value \( m \)

1. **Using the AM-GM Inequality**:
\[ \frac{3^{y_1} + 3^{y_2} + 3^{y_3}}{3} \geq \sqrt[3]{3^{y_1} \cdot 3^{y_2} \cdot 3^{y_3}} \]
This implies:
\[ 3^{y_1} + 3^{y_2} + 3^{y_3} \geq 3 \sqrt[3]{3^{y_1 + y_2 + y_3}} = 3 \sqrt[3]{3^9} = 3 \cdot 27 = 81 \]

2. **Taking the logarithm**:
\[ \log_3(3^{y_1} + 3^{y_2} + 3^{y_3}) \geq \log_3(81) = \log_3(3^4) = 4 \]
Thus, the minimum value \( m = 4 \).

### Step 2: Finding the maximum value \( M \)

1. **Using the properties of logarithms**:
\[ \log_3(x_1) + \log_3(x_2) + \log_3(x_3) = \log_3(x_1 \cdot x_2 \cdot x_3) \]

2. **Using the AM-GM Inequality again**:
\[ \frac{x_1 + x_2 + x_3}{3} \geq \sqrt[3]{x_1 x_2 x_3} \]
This implies:
\[ 3 \geq \sqrt[3]{x_1 x_2 x_3} \implies x_1 x_2 x_3 \leq 27 \]

3. **Taking the logarithm**:
\[ \log_3(x_1 x_2 x_3) \leq \log_3(27) = \log_3(3^3) = 3 \]
Thus, the maximum value \( M = 3 \).

### Step 3: Calculating \( \log_2(m^3) + \log_3(M^2) \)

1. **Calculating \( m^3 \)**:
\[ m^3 = 4^3 = 64 \implies \log_2(m^3) = \log_2(64) = 6 \]

2. **Calculating \( M^2 \)**:
\[ M^2 = 3^2 = 9 \implies \log_3(M^2) = \log_3(9) = 2 \]

3. **Final Calculation**:
\[ \log_2(m^3) + \log_3(M^2) = 6 + 2 = 8 \]

### Final Answer
The value of \( \log_2(m^3) + \log_3(M^2) \) is \( \boxed{8} \).