Express with rational denominator :
`(4)/(root3((9))-root3(3)+1)`

Answer

To express the fraction \(\frac{4}{\sqrt[3]{9} - \sqrt[3]{3} + 1}\) with a rational denominator, we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ n = \frac{4}{\sqrt[3]{9} - \sqrt[3]{3} + 1} \]

### Step 2: Simplify the cube roots
We can rewrite \(\sqrt[3]{9}\) as \(\sqrt[3]{3^2} = 3^{2/3}\) and \(\sqrt[3]{3}\) as \(3^{1/3}\):
\[ n = \frac{4}{3^{2/3} - 3^{1/3} + 1} \]

### Step 3: Rationalize the denominator
To rationalize the denominator, we can multiply the numerator and denominator by the conjugate of the denominator. The conjugate in this case can be expressed as:
\[ (3^{1/3} + 1)^2 - (3^{1/3})^2 \]
This is based on the identity \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\), where \(a = 3^{1/3}\) and \(b = 1\).

### Step 4: Multiply numerator and denominator
Now, we multiply both the numerator and denominator by \((\sqrt[3]{3} + 1)^2\):
\[ n = \frac{4(\sqrt[3]{3} + 1)^2}{(\sqrt[3]{9} - \sqrt[3]{3} + 1)(\sqrt[3]{3} + 1)^2} \]

### Step 5: Expand the denominator
Now we need to expand the denominator:
\[ (\sqrt[3]{9} - \sqrt[3]{3} + 1)(\sqrt[3]{3} + 1)^2 = (\sqrt[3]{9} - \sqrt[3]{3} + 1)(\sqrt[3]{9} + 2\sqrt[3]{3} + 1) \]
This will simplify to:
\[ = \sqrt[3]{27} - \sqrt[3]{9} + \sqrt[3]{9} + 2\sqrt[3]{3} + 1 = 3 + 2\sqrt[3]{3} + 1 = 4 + 2\sqrt[3]{3} \]

### Step 6: Final expression
Now we have:
\[ n = \frac{4(\sqrt[3]{3} + 1)^2}{4 + 2\sqrt[3]{3}} \]
We can simplify this further if needed, but this is the expression with a rational denominator.