A rationalising factor of `(root3(9)-root3(3)+1)` is

Options

A. `root3(3)-1`
B. `root3(3)+1`
C. `root3(9)+1`
D. `root3(9)-1`

Answer

To find the rationalizing factor of the expression \( \sqrt[3]{9} - \sqrt[3]{3} + 1 \), we will follow these steps:

### Step 1: Rewrite the expression in terms of exponents
We can express the cube roots in terms of exponents:
\[ \sqrt[3]{9} = 9^{1/3} = (3^2)^{1/3} = 3^{2/3} \]
\[ \sqrt[3]{3} = 3^{1/3} \]
Thus, our expression becomes:
\[ 3^{2/3} - 3^{1/3} + 1 \]

### Step 2: Identify the terms for the rationalizing factor
We can denote:
- \( A = 3^{1/3} \)
- \( B = 1 \)

The expression can now be rewritten as:
\[ A^2 - A + B \]

### Step 3: Use the formula for cube roots
We know that the rationalizing factor can be derived from the identity for the sum of cubes:
\[ A^3 + B^3 = (A + B)(A^2 - AB + B^2) \]
Here, we have:
- \( A^3 = (3^{1/3})^3 = 3 \)
- \( B^3 = 1^3 = 1 \)
Thus:
\[ A^3 + B^3 = 3 + 1 = 4 \]

### Step 4: Find the rationalizing factor
To rationalize the expression \( A^2 - A + B \), we multiply it by \( A + B \):
\[ (A - B)(A^2 - AB + B^2) = (3^{1/3} + 1)(3^{2/3} - 3^{1/3} + 1) \]

### Step 5: Write the final answer
The rationalizing factor of the expression \( \sqrt[3]{9} - \sqrt[3]{3} + 1 \) is:
\[ \sqrt[3]{3} + 1 \]