If `log_(b) a. log_(c ) a + log_(a) b. log_(c ) b + log_(a) c. log_(b) c = 3` (where a, b, c are different positive real number `ne 1`), then find the value of a b c.

Answer

To solve the equation \( \log_b a \cdot \log_c a + \log_a b \cdot \log_c b + \log_a c \cdot \log_b c = 3 \) where \( a, b, c \) are different positive real numbers not equal to 1, we will follow these steps:

### Step 1: Rewrite the logarithms using natural logarithms
We can use the change of base formula for logarithms, which states that:
\[ \log_b a = \frac{\ln a}{\ln b} \]
Therefore, we can rewrite the equation as:
\[ \frac{\ln a}{\ln b} \cdot \frac{\ln a}{\ln c} + \frac{\ln b}{\ln a} \cdot \frac{\ln b}{\ln c} + \frac{\ln c}{\ln a} \cdot \frac{\ln c}{\ln b} = 3 \]

### Step 2: Simplify the equation
This can be simplified to:
\[ \frac{\ln^2 a}{\ln b \cdot \ln c} + \frac{\ln^2 b}{\ln a \cdot \ln c} + \frac{\ln^2 c}{\ln a \cdot \ln b} = 3 \]
Now, we can take the common denominator, which is \( \ln a \cdot \ln b \cdot \ln c \):
\[ \frac{\ln^2 a \cdot \ln a + \ln^2 b \cdot \ln b + \ln^2 c \cdot \ln c}{\ln a \cdot \ln b \cdot \ln c} = 3 \]

### Step 3: Multiply both sides by the denominator
Multiplying both sides by \( \ln a \cdot \ln b \cdot \ln c \) gives us:
\[ \ln^2 a \cdot \ln c + \ln^2 b \cdot \ln c + \ln^2 c \cdot \ln b = 3 \cdot \ln a \cdot \ln b \cdot \ln c \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ \ln^2 a + \ln^2 b + \ln^2 c = 3 \cdot \ln a \cdot \ln b \cdot \ln c \]

### Step 5: Recognizing the identity
This equation resembles the identity for sums of cubes:
\[ x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - xz - yz) \]
If we let \( x = \ln a \), \( y = \ln b \), and \( z = \ln c \), we can conclude that:
\[ \ln a + \ln b + \ln c = 0 \]

### Step 6: Exponentiating both sides
Exponentiating both sides gives:
\[ e^{\ln a + \ln b + \ln c} = e^0 \]
This simplifies to:
\[ abc = 1 \]

### Final Answer
Thus, the value of \( abc \) is:
\[ \boxed{1} \]