Given that `log_(x)y,log_(z)x,log_(y)z` are in GP, xyz=64 and `x^(3),y^(3),z^(3)` are in AP.
Q. Which one of the following is correct ? x,y and z are

Options

A. In AP only
B. In GP only
C. In both AP and GP
D. Neither in AP nor in GP

Answer

To solve the problem, we will follow the steps outlined in the video transcript, ensuring that we derive the necessary relationships and conditions from the given information.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
We are given that \( \log_{x} y, \log_{z} x, \log_{y} z \) are in geometric progression (GP). Additionally, we know that \( xyz = 64 \) and \( x^3, y^3, z^3 \) are in arithmetic progression (AP).

2. **Using the GP Condition**:
For three terms \( a, b, c \) to be in GP, the relationship \( b^2 = ac \) must hold. Here, we can let:
- \( a = \log_{x} y \)
- \( b = \log_{z} x \)
- \( c = \log_{y} z \)

Therefore, we can write:
\[ (\log_{z} x)^2 = \log_{x} y \cdot \log_{y} z \]

3. **Applying Logarithmic Properties**:
Using the change of base formula, we can express the logarithms in terms of natural logarithms:
\[ \log_{x} y = \frac{\log y}{\log x}, \quad \log_{z} x = \frac{\log x}{\log z}, \quad \log_{y} z = \frac{\log z}{\log y} \]
Substituting these into our GP condition gives:
\[ \left(\frac{\log x}{\log z}\right)^2 = \left(\frac{\log y}{\log x}\right) \left(\frac{\log z}{\log y}\right) \]
Simplifying the right side:
\[ \left(\frac{\log x}{\log z}\right)^2 = \frac{\log z}{\log x} \]
Cross-multiplying yields:
\[ (\log x)^3 = (\log z)^2 \]

4. **Using the Product Condition**:
We know \( xyz = 64 \). Taking logarithms gives:
\[ \log(xyz) = \log 64 \implies \log x + \log y + \log z = 6 \log 2 \]

5. **Using the AP Condition**:
Since \( x^3, y^3, z^3 \) are in AP, we have:
\[ 2y^3 = x^3 + z^3 \]
Using the identity for cubes:
\[ 2y^3 = (x + z)(x^2 - xz + z^2) \]

6. **Substituting Relationships**:
From the GP condition, we found \( x = z \). Let \( x = z = k \). Then:
\[ k^2y = 64 \implies y = \frac{64}{k^2} \]
Substituting into the AP condition:
\[ 2\left(\frac{64}{k^2}\right)^3 = k^3 + k^3 \implies 2\left(\frac{64^3}{k^6}\right) = 2k^3 \]
Simplifying gives:
\[ \frac{64^3}{k^6} = k^3 \implies 64^3 = k^9 \implies k = 4 \]

7. **Finding Values of x, y, z**:
Since \( k = 4 \):
\[ x = 4, \quad y = \frac{64}{4^2} = 4, \quad z = 4 \]

8. **Conclusion**:
Thus, \( x = y = z = 4 \). Since all three values are equal, they are in both arithmetic progression (AP) and geometric progression (GP).

### Final Answer:
The correct option is that \( x, y, z \) are in both AP and GP.