If x = `sqrt(3)- sqrt(2)` find the value of
`x^(3)+ (1)/(x^(3))`

Answer

To find the value of \( x^3 + \frac{1}{x^3} \) where \( x = \sqrt{3} - \sqrt{2} \), we will follow these steps:

### Step 1: Calculate \( x^3 \)
Using the formula for the cube of a binomial, we have:
\[ x^3 = (\sqrt{3} - \sqrt{2})^3 = (\sqrt{3})^3 - 3(\sqrt{3})^2(\sqrt{2}) + 3(\sqrt{3})(\sqrt{2})^2 - (\sqrt{2})^3 \]
Calculating each term:
- \( (\sqrt{3})^3 = 3\sqrt{3} \)
- \( 3(\sqrt{3})^2(\sqrt{2}) = 3 \cdot 3 \cdot \sqrt{2} = 9\sqrt{2} \)
- \( 3(\sqrt{3})(\sqrt{2})^2 = 3\sqrt{3} \cdot 2 = 6\sqrt{3} \)
- \( (\sqrt{2})^3 = 2\sqrt{2} \)

Putting it all together:
\[ x^3 = 3\sqrt{3} - 9\sqrt{2} + 6\sqrt{3} - 2\sqrt{2} \]
Combining like terms:
\[ x^3 = (3\sqrt{3} + 6\sqrt{3}) + (-9\sqrt{2} - 2\sqrt{2}) = 9\sqrt{3} - 11\sqrt{2} \]

### Step 2: Calculate \( \frac{1}{x} \)
To find \( \frac{1}{x} \), we rationalize the denominator:
\[ \frac{1}{x} = \frac{1}{\sqrt{3} - \sqrt{2}} \cdot \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = \frac{\sqrt{3} + \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2} = \frac{\sqrt{3} + \sqrt{2}}{3 - 2} = \sqrt{3} + \sqrt{2} \]

### Step 3: Calculate \( \left( \frac{1}{x} \right)^3 \)
Now we need to find \( \left( \sqrt{3} + \sqrt{2} \right)^3 \):
\[ \left( \sqrt{3} + \sqrt{2} \right)^3 = (\sqrt{3})^3 + 3(\sqrt{3})^2(\sqrt{2}) + 3(\sqrt{3})(\sqrt{2})^2 + (\sqrt{2})^3 \]
Calculating each term:
- \( (\sqrt{3})^3 = 3\sqrt{3} \)
- \( 3(\sqrt{3})^2(\sqrt{2}) = 9\sqrt{2} \)
- \( 3(\sqrt{3})(\sqrt{2})^2 = 6\sqrt{3} \)
- \( (\sqrt{2})^3 = 2\sqrt{2} \)

Putting it all together:
\[ \left( \frac{1}{x} \right)^3 = 3\sqrt{3} + 9\sqrt{2} + 6\sqrt{3} + 2\sqrt{2} \]
Combining like terms:
\[ \left( \frac{1}{x} \right)^3 = (3\sqrt{3} + 6\sqrt{3}) + (9\sqrt{2} + 2\sqrt{2}) = 9\sqrt{3} + 11\sqrt{2} \]

### Step 4: Calculate \( x^3 + \frac{1}{x^3} \)
Now we can find \( x^3 + \frac{1}{x^3} \):
\[ x^3 + \frac{1}{x^3} = (9\sqrt{3} - 11\sqrt{2}) + (9\sqrt{3} + 11\sqrt{2}) = 18\sqrt{3} \]

### Final Answer:
\[ x^3 + \frac{1}{x^3} = 18\sqrt{3} \]