If `log_(a)b+log_(b)c+log_(c)a` vanishes where a, b and c are positive reals different from unity then the value of `(log_(a)b)^(3) + (log_(b)c)^(3) + (log_(c)a)^(3)` is

Options

A. an odd prime
B. an even prime
C. an odd composite
D. an irrational number

Answer

To solve the problem, we need to find the value of \((\log_a b)^3 + (\log_b c)^3 + (\log_c a)^3\) given that \(\log_a b + \log_b c + \log_c a = 0\).

### Step-by-Step Solution:

1. **Define Variables**:
Let \(x = \log_a b\), \(y = \log_b c\), and \(z = \log_c a\). According to the problem, we have:
\[ x + y + z = 0 \]

2. **Use the Identity for Cubes**:
We can use the identity for the sum of cubes:
\[ x^3 + y^3 + z^3 = 3xyz \]
This identity holds true when \(x + y + z = 0\).

3. **Express the Required Value**:
We need to find \(x^3 + y^3 + z^3\). From the identity, we can substitute:
\[ x^3 + y^3 + z^3 = 3xyz \]

4. **Calculate \(xyz\)**:
Now we need to find the value of \(xyz\):
\[ xyz = \log_a b \cdot \log_b c \cdot \log_c a \]
Using the change of base formula, we can express each logarithm:
\[ \log_a b = \frac{\log b}{\log a}, \quad \log_b c = \frac{\log c}{\log b}, \quad \log_c a = \frac{\log a}{\log c} \]

5. **Substituting into \(xyz\)**:
Substitute these into the expression for \(xyz\):
\[ xyz = \left(\frac{\log b}{\log a}\right) \left(\frac{\log c}{\log b}\right) \left(\frac{\log a}{\log c}\right) \]
Notice that \(\log b\) in the numerator and denominator cancels, as does \(\log c\), leading to:
\[ xyz = 1 \]

6. **Final Calculation**:
Now substituting \(xyz = 1\) back into the equation for \(x^3 + y^3 + z^3\):
\[ x^3 + y^3 + z^3 = 3 \cdot 1 = 3 \]

### Conclusion:
Thus, the value of \((\log_a b)^3 + (\log_b c)^3 + (\log_c a)^3\) is:
\[ \boxed{3} \]