If 3a + 4b =9 and ab= 2 find the value of :` 27 a^(3) + 64 b ^(3)`

Answer

To solve the problem, we need to find the value of \( 27a^3 + 64b^3 \) given the equations \( 3a + 4b = 9 \) and \( ab = 2 \).

### Step-by-Step Solution:

1. **Start with the given equations:**
\[ 3a + 4b = 9 \quad \text{(1)} \]
\[ ab = 2 \quad \text{(2)} \]

2. **Cube both sides of equation (1):**
\[ (3a + 4b)^3 = 9^3 \]
\[ (3a + 4b)^3 = 729 \]

3. **Use the identity for the cube of a binomial:**
\[ (x + y)^3 = x^3 + y^3 + 3xy(x + y) \]
Here, let \( x = 3a \) and \( y = 4b \):
\[ (3a + 4b)^3 = (3a)^3 + (4b)^3 + 3(3a)(4b)(3a + 4b) \]

4. **Substituting the values:**
\[ 729 = 27a^3 + 64b^3 + 3(12ab)(9) \]
Since \( ab = 2 \):
\[ 729 = 27a^3 + 64b^3 + 3(12 \cdot 2)(9) \]
\[ 729 = 27a^3 + 64b^3 + 3(24)(9) \]
\[ 729 = 27a^3 + 64b^3 + 648 \]

5. **Isolate \( 27a^3 + 64b^3 \):**
\[ 27a^3 + 64b^3 = 729 - 648 \]
\[ 27a^3 + 64b^3 = 81 \]

### Final Answer:
\[ 27a^3 + 64b^3 = 81 \]