If `x = a + (1)/(a)` and `y = a - (1)/(a)` then `sqrt(x^(4) + y^(4) - 2x^(2)y^(2))` is equal to :

Options

A. `16a^(2)`
B. 8
C. `(8)/(a^(2))`
D. 4

Answer

To solve the problem, we need to evaluate the expression \( \sqrt{x^4 + y^4 - 2x^2y^2} \) given that \( x = a + \frac{1}{a} \) and \( y = a - \frac{1}{a} \).

### Step-by-Step Solution:

1. **Substituting the values of \( x \) and \( y \)**:
\[ x = a + \frac{1}{a}, \quad y = a - \frac{1}{a} \]

2. **Finding \( x^2 \) and \( y^2 \)**:
\[ x^2 = \left(a + \frac{1}{a}\right)^2 = a^2 + 2 + \frac{1}{a^2} \]
\[ y^2 = \left(a - \frac{1}{a}\right)^2 = a^2 - 2 + \frac{1}{a^2} \]

3. **Calculating \( x^2y^2 \)**:
\[ x^2y^2 = \left(a^2 + 2 + \frac{1}{a^2}\right)\left(a^2 - 2 + \frac{1}{a^2}\right) \]
Using the difference of squares:
\[ x^2y^2 = (a^2 + \frac{1}{a^2})^2 - 2^2 = (a^2 + \frac{1}{a^2})^2 - 4 \]

4. **Finding \( x^4 + y^4 \)**:
\[ x^4 + y^4 = (x^2)^2 + (y^2)^2 \]
Using the identity \( a^2 + b^2 = (a + b)^2 - 2ab \):
\[ x^4 + y^4 = (x^2 + y^2)^2 - 2x^2y^2 \]

5. **Calculating \( x^2 + y^2 \)**:
\[ x^2 + y^2 = \left(a^2 + 2 + \frac{1}{a^2}\right) + \left(a^2 - 2 + \frac{1}{a^2}\right) = 2a^2 + 2\frac{1}{a^2} \]

6. **Putting it all together**:
\[ \sqrt{x^4 + y^4 - 2x^2y^2} = \sqrt{(x^2 + y^2)^2 - 2x^2y^2} \]
This simplifies to:
\[ \sqrt{(x^2 - y^2)^2} \]
Therefore:
\[ \sqrt{x^4 + y^4 - 2x^2y^2} = |x^2 - y^2| \]

7. **Finding \( x^2 - y^2 \)**:
\[ x^2 - y^2 = (a^2 + 2 + \frac{1}{a^2}) - (a^2 - 2 + \frac{1}{a^2}) = 4 \]

8. **Final Result**:
\[ \sqrt{x^4 + y^4 - 2x^2y^2} = |4| = 4 \]

### Conclusion:
The final answer is \( 4 \).