The rationalising factor of `root(7)(x^(3)y^(5)z^(2))` is _______.

Options

A. `root(7)(z^(3)y^(2)x^(5))`
B. `root(4)(x^(3)y^(2)z)`
C. `sqrt(x^(4)y^(2)z^(5))`
D. `root(7)(y^(2)x^(4)z^(5))`

Answer

To find the rationalizing factor of \( \sqrt[7]{x^3 y^5 z^2} \), we need to ensure that when we multiply it by the rationalizing factor, the resulting expression is a perfect seventh power.

### Step-by-Step Solution:

1. **Identify the expression**: We have \( \sqrt[7]{x^3 y^5 z^2} \).

2. **Determine the powers**:
- For \( x \): The power is 3.
- For \( y \): The power is 5.
- For \( z \): The power is 2.

3. **Calculate the required powers to make them multiples of 7**:
- For \( x^3 \): To make it a multiple of 7, we need \( 7 - 3 = 4 \). So, we need \( x^4 \).
- For \( y^5 \): To make it a multiple of 7, we need \( 7 - 5 = 2 \). So, we need \( y^2 \).
- For \( z^2 \): To make it a multiple of 7, we need \( 7 - 2 = 5 \). So, we need \( z^5 \).

4. **Combine the rationalizing factors**:
- The rationalizing factor will be \( \sqrt[7]{x^4 y^2 z^5} \).

5. **Final answer**: The rationalizing factor of \( \sqrt[7]{x^3 y^5 z^2} \) is \( \sqrt[7]{x^4 y^2 z^5} \).