Statement-1: If `a =y^(2), b=z^(2) " and " c= x^(2), " then log"_(a) x^(3) xx "log"_(b) y^(3) xx "log"_(c)z^(3) = (27)/(8)`
Statement-2: `"log"_(b) a = (1)/("log"_(a)b)`

Options

A. Statement-1 is True, Statement-2 is true, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is true, Statement-2 is NOT a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To solve the given problem, we will analyze Statement-1 and verify its correctness step by step.

### Step 1: Understand the given variables
We have:
- \( a = y^2 \)
- \( b = z^2 \)
- \( c = x^2 \)

### Step 2: Rewrite the logarithmic expressions
We need to evaluate:
\[ \log_a (x^3) \cdot \log_b (y^3) \cdot \log_c (z^3) \]

Using the change of base formula for logarithms, we can rewrite each logarithm:
\[ \log_a (x^3) = \frac{\log(x^3)}{\log(a)} = \frac{3 \log(x)}{\log(y^2)} = \frac{3 \log(x)}{2 \log(y)} \]
\[ \log_b (y^3) = \frac{\log(y^3)}{\log(b)} = \frac{3 \log(y)}{\log(z^2)} = \frac{3 \log(y)}{2 \log(z)} \]
\[ \log_c (z^3) = \frac{\log(z^3)}{\log(c)} = \frac{3 \log(z)}{\log(x^2)} = \frac{3 \log(z)}{2 \log(x)} \]

### Step 3: Substitute back into the expression
Now substituting these back into the original expression:
\[ \log_a (x^3) \cdot \log_b (y^3) \cdot \log_c (z^3) = \left(\frac{3 \log(x)}{2 \log(y)}\right) \cdot \left(\frac{3 \log(y)}{2 \log(z)}\right) \cdot \left(\frac{3 \log(z)}{2 \log(x)}\right) \]

### Step 4: Simplify the expression
Multiplying these together:
\[ = \frac{3 \log(x)}{2 \log(y)} \cdot \frac{3 \log(y)}{2 \log(z)} \cdot \frac{3 \log(z)}{2 \log(x)} \]
Notice that \( \log(x) \), \( \log(y) \), and \( \log(z) \) will cancel out:
\[ = \frac{3 \cdot 3 \cdot 3}{2 \cdot 2 \cdot 2} = \frac{27}{8} \]

### Conclusion
Thus, we have shown that:
\[ \log_a (x^3) \cdot \log_b (y^3) \cdot \log_c (z^3) = \frac{27}{8} \]
This confirms that Statement-1 is true.

### Step 5: Verify Statement-2
Statement-2 states:
\[ \log_b a = \frac{1}{\log_a b} \]
This is indeed a property of logarithms and is true. However, it was not used in the proof of Statement-1.

### Final Answer
- Statement-1 is true.
- Statement-2 is true but does not explain Statement-1.