If `3` rational numbers `x^(1//x)` , `y^(1//y)` and `z^(1//z)` are equal and `x^(yz)+y^(zx)+z^(xy)=729` , then find the value of `x^(1//x)`

Options

A. `6`
B. `3`
C. `5`
D. `4`

Explanation

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
We have three rational numbers \( x^{1/x} \), \( y^{1/y} \), and \( z^{1/z} \) that are equal. Let's denote this common value as \( k \).
\[ x^{1/x} = k, \quad y^{1/y} = k, \quad z^{1/z} = k \]

2. **Expressing \( x \), \( y \), and \( z \) in Terms of \( k \)**:
From the equation \( x^{1/x} = k \), we can express \( x \) as:
\[ x = k^x \]
Similarly, we can express \( y \) and \( z \):
\[ y = k^y, \quad z = k^z \]

3. **Substituting into the Given Equation**:
We are given the equation:
\[ x^{yz} + y^{zx} + z^{xy} = 729 \]
Substituting \( x \), \( y \), and \( z \) in terms of \( k \):
\[ (k^x)^{yz} + (k^y)^{zx} + (k^z)^{xy} = 729 \]

4. **Simplifying the Terms**:
This can be simplified using the properties of exponents:
\[ k^{xyz} + k^{xyz} + k^{xyz} = 729 \]
Since all three terms are equal, we can combine them:
\[ 3k^{xyz} = 729 \]

5. **Solving for \( k^{xyz} \)**:
Dividing both sides by 3:
\[ k^{xyz} = \frac{729}{3} = 243 \]

6. **Expressing 243 as a Power of 3**:
We know that:
\[ 243 = 3^5 \]
Therefore, we have:
\[ k^{xyz} = 3^5 \]

7. **Finding the Value of \( k \)**:
Since \( k^{xyz} = 3^5 \), we can equate the bases:
\[ k = 3 \]

8. **Finding \( x^{1/x} \)**:
We need to find the value of \( x^{1/x} \), which we defined as \( k \):
\[ x^{1/x} = k = 3 \]

### Final Answer:
Thus, the value of \( x^{1/x} \) is \( \boxed{3} \).