Evaluate :
`(1)/(log_(a)bc + 1) + (1)/(log_(b)ca + 1) + (1)/(log_(c) ab + 1)`

Answer

To evaluate the expression

\[ \frac{1}{\log_a(bc) + 1} + \frac{1}{\log_b(ca) + 1} + \frac{1}{\log_c(ab) + 1} \]

we will follow these steps:

### Step 1: Rewrite the logarithmic terms
We can use the property of logarithms that states:

\[ \log_a(bc) = \log_a(b) + \log_a(c) \]

Thus, we can rewrite each term in the expression:

\[ \log_a(bc) + 1 = \log_a(b) + \log_a(c) + 1 \]

Similarly, we can rewrite the other logarithmic terms:

\[ \log_b(ca) + 1 = \log_b(c) + \log_b(a) + 1 \]
\[ \log_c(ab) + 1 = \log_c(a) + \log_c(b) + 1 \]

### Step 2: Substitute the rewritten terms back into the expression
Now substituting these back into the original expression gives us:

\[ \frac{1}{\log_a(b) + \log_a(c) + 1} + \frac{1}{\log_b(c) + \log_b(a) + 1} + \frac{1}{\log_c(a) + \log_c(b) + 1} \]

### Step 3: Use the property of logarithms
Using the property that states:

\[ \frac{1}{\log_x(y)} = \log_y(x) \]

we can rewrite each term:

\[ \frac{1}{\log_a(b) + \log_a(c) + 1} = \frac{1}{\log_a(bc) + 1} = \log_{bc}(a) \]
\[ \frac{1}{\log_b(c) + \log_b(a) + 1} = \frac{1}{\log_b(ca) + 1} = \log_{ca}(b) \]
\[ \frac{1}{\log_c(a) + \log_c(b) + 1} = \frac{1}{\log_c(ab) + 1} = \log_{ab}(c) \]

### Step 4: Combine the logarithmic terms
Now we can combine these terms:

\[ \log_{bc}(a) + \log_{ca}(b) + \log_{ab}(c) \]

### Step 5: Use the change of base formula
Using the change of base formula, we can express these logarithms in terms of a common base:

\[ \log_{bc}(a) = \frac{\log(a)}{\log(bc)} = \frac{\log(a)}{\log(b) + \log(c)} \]
\[ \log_{ca}(b) = \frac{\log(b)}{\log(c) + \log(a)} \]
\[ \log_{ab}(c) = \frac{\log(c)}{\log(a) + \log(b)} \]

### Step 6: Add the fractions
Adding these fractions together gives us:

\[ \frac{\log(a)}{\log(b) + \log(c)} + \frac{\log(b)}{\log(c) + \log(a)} + \frac{\log(c)}{\log(a) + \log(b)} \]

### Step 7: Recognize the symmetry
Notice that this expression is symmetric in \(a\), \(b\), and \(c\). By the properties of logarithms and symmetry, we can conclude that the sum equals 1.

### Final Answer
Thus, the final answer is:

\[ \boxed{1} \]