If `x = 3 + 2 sqrt2,` then the value of `x ^(3) + (1)/(x ^(3)) and x ^(3) - (1)/(x ^(3))` ae respectively:

Options

A. 234216
B. 216234
C. `198,140 sqrt2`
D. `140 sqrt2, 198`

Answer

To solve the problem, we need to find the values of \( x^3 + \frac{1}{x^3} \) and \( x^3 - \frac{1}{x^3} \) given that \( x = 3 + 2\sqrt{2} \).

### Step 1: Find \( \frac{1}{x} \)
First, we will find \( \frac{1}{x} \) by rationalizing it.

\[ \frac{1}{x} = \frac{1}{3 + 2\sqrt{2}} \]

To rationalize, we multiply the numerator and the denominator by the conjugate of the denominator:

\[ \frac{1}{x} = \frac{1 \cdot (3 - 2\sqrt{2})}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})} \]

Calculating the denominator:

\[ (3 + 2\sqrt{2})(3 - 2\sqrt{2}) = 3^2 - (2\sqrt{2})^2 = 9 - 8 = 1 \]

Thus,

\[ \frac{1}{x} = 3 - 2\sqrt{2} \]

### Step 2: Find \( x + \frac{1}{x} \)
Now we can find \( x + \frac{1}{x} \):

\[ x + \frac{1}{x} = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6 \]

### Step 3: Find \( x^3 + \frac{1}{x^3} \)
Using the identity:

\[ x^3 + \frac{1}{x^3} = \left( x + \frac{1}{x} \right)^3 - 3 \left( x + \frac{1}{x} \right) \]

Substituting \( x + \frac{1}{x} = 6 \):

\[ x^3 + \frac{1}{x^3} = 6^3 - 3 \cdot 6 \]

Calculating \( 6^3 \):

\[ 6^3 = 216 \]

Now substituting back:

\[ x^3 + \frac{1}{x^3} = 216 - 18 = 198 \]

### Step 4: Find \( x - \frac{1}{x} \)
Next, we find \( x - \frac{1}{x} \):

\[ x - \frac{1}{x} = (3 + 2\sqrt{2}) - (3 - 2\sqrt{2}) = 4\sqrt{2} \]

### Step 5: Find \( x^3 - \frac{1}{x^3} \)
Using the identity:

\[ x^3 - \frac{1}{x^3} = \left( x - \frac{1}{x} \right)^3 + 3 \left( x - \frac{1}{x} \right) \]

Substituting \( x - \frac{1}{x} = 4\sqrt{2} \):

\[ x^3 - \frac{1}{x^3} = (4\sqrt{2})^3 + 3(4\sqrt{2}) \]

Calculating \( (4\sqrt{2})^3 \):

\[ (4\sqrt{2})^3 = 64 \cdot 2\sqrt{2} = 128\sqrt{2} \]

Now substituting back:

\[ x^3 - \frac{1}{x^3} = 128\sqrt{2} + 12\sqrt{2} = 140\sqrt{2} \]

### Final Answer
The values are:

\[ x^3 + \frac{1}{x^3} = 198 \]
\[ x^3 - \frac{1}{x^3} = 140\sqrt{2} \]