If `x = 1/(sqrt(2) - 1)`, then the value of `x^2 - 6 + 1/(x^2)` is :

Options

A. 0
B. 1
C. `sqrt(2)`
D. can't be determined

Answer

To solve the problem, we need to find the value of \( x^2 - 6 + \frac{1}{x^2} \) given that \( x = \frac{1}{\sqrt{2} - 1} \).

### Step 1: Rationalize \( x \)
We start with the expression for \( x \):
\[ x = \frac{1}{\sqrt{2} - 1} \]
To rationalize the denominator, we multiply the numerator and the denominator by \( \sqrt{2} + 1 \):
\[ x = \frac{1 \cdot (\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{\sqrt{2} + 1}{2 - 1} = \sqrt{2} + 1 \]

### Step 2: Find \( \frac{1}{x} \)
Next, we find \( \frac{1}{x} \):
\[ \frac{1}{x} = \frac{1}{\sqrt{2} + 1} \]
Again, we rationalize the denominator:
\[ \frac{1}{x} = \frac{1 \cdot (\sqrt{2} - 1)}{(\sqrt{2} + 1)(\sqrt{2} - 1)} = \frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1 \]

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

### Step 4: Find \( x^2 + \frac{1}{x^2} \)
Using the identity \( (a + b)^2 = a^2 + b^2 + 2ab \), we can find \( x^2 + \frac{1}{x^2} \):
\[ (x + \frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2} \]
Substituting \( x + \frac{1}{x} = 2\sqrt{2} \):
\[ (2\sqrt{2})^2 = x^2 + 2 + \frac{1}{x^2} \]
\[ 8 = x^2 + 2 + \frac{1}{x^2} \]
Thus,
\[ x^2 + \frac{1}{x^2} = 8 - 2 = 6 \]

### Step 5: Calculate \( x^2 - 6 + \frac{1}{x^2} \)
Now we substitute \( x^2 + \frac{1}{x^2} \) into the expression we need to evaluate:
\[ x^2 - 6 + \frac{1}{x^2} = 6 - 6 = 0 \]

### Final Answer
The value of \( x^2 - 6 + \frac{1}{x^2} \) is:
\[ \boxed{0} \]