If `log_(a)b+log_(b)c+log_(c)a` vanishes where a, b and c are positive reals different than 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 1: Set up the equation
Let:
- \(x = \log_a b\)
- \(y = \log_b c\)
- \(z = \log_c a\)

From the problem, we know:
\[ x + y + z = 0 \]

### Step 2: Use the identity for cubes
We use the identity for the sum of cubes:
\[ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \]
Since \(x + y + z = 0\), we can simplify this to:
\[ x^3 + y^3 + z^3 = 3xyz \]

### Step 3: Calculate \(xyz\)
Now, we need to find \(xyz\):
\[ xyz = \log_a b \cdot \log_b c \cdot \log_c a \]
Using the change of base formula, we can express this as:
\[ xyz = \frac{\log b}{\log a} \cdot \frac{\log c}{\log b} \cdot \frac{\log a}{\log c} \]
Notice that the terms \(\log b\), \(\log c\), and \(\log a\) will cancel out:
\[ xyz = 1 \]

### Step 4: Substitute back into the equation
Now substituting \(xyz = 1\) back into the equation for the sum of cubes:
\[ x^3 + y^3 + z^3 = 3xyz = 3 \cdot 1 = 3 \]

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