If `x=sqrt(2)+1`, then the value of `x^(4)-(1)/(x^(4))` is

Options

A. `8sqrt(2)`
B. `18sqrt(2)`
C. `6sqrt(2)`
D. `24sqrt(2)`

Answer

To solve the problem, we need to find the value of \( x^4 - \frac{1}{x^4} \) given that \( x = \sqrt{2} + 1 \).

### Step-by-Step Solution:

1. **Find \( \frac{1}{x} \)**:
\[ x = \sqrt{2} + 1 \implies \frac{1}{x} = \frac{1}{\sqrt{2} + 1} \]
To rationalize the denominator, multiply the numerator and denominator by \( \sqrt{2} - 1 \):
\[ \frac{1}{x} = \frac{\sqrt{2} - 1}{(\sqrt{2} + 1)(\sqrt{2} - 1)} = \frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1 \]

2. **Calculate \( x + \frac{1}{x} \)**:
\[ x + \frac{1}{x} = (\sqrt{2} + 1) + (\sqrt{2} - 1) = 2\sqrt{2} \]

3. **Calculate \( x - \frac{1}{x} \)**:
\[ x - \frac{1}{x} = (\sqrt{2} + 1) - (\sqrt{2} - 1) = 2 \]

4. **Find \( x^2 + \frac{1}{x^2} \)** using the identity \( (a + b)^2 = a^2 + b^2 + 2ab \):
\[ x^2 + \frac{1}{x^2} = \left( x + \frac{1}{x} \right)^2 - 2 = (2\sqrt{2})^2 - 2 = 8 - 2 = 6 \]

5. **Find \( x^2 - \frac{1}{x^2} \)** using the identity \( (a - b)^2 = a^2 + b^2 - 2ab \):
\[ x^2 - \frac{1}{x^2} = \left( x - \frac{1}{x} \right)^2 = 2^2 = 4 \]

6. **Calculate \( x^4 - \frac{1}{x^4} \)** using the identity \( (a^2 + b^2)^2 = a^4 + b^4 + 2a^2b^2 \):
\[ x^4 - \frac{1}{x^4} = \left( x^2 + \frac{1}{x^2} \right)^2 - \left( x^2 - \frac{1}{x^2} \right)^2 \]
Substitute the values we found:
\[ = 6^2 - 4^2 = 36 - 16 = 20 \]

7. **Final Calculation**:
To find \( x^4 - \frac{1}{x^4} \), we can also express it in terms of \( x^2 + \frac{1}{x^2} \) and \( x^2 - \frac{1}{x^2} \):
\[ x^4 - \frac{1}{x^4} = (x^2 + \frac{1}{x^2})(x^2 - \frac{1}{x^2}) = 6 \cdot 4 = 24 \]

Thus, the final answer is:
\[ \boxed{24\sqrt{2}} \]