If `x = a ^((1)/(2) ) + a ^((-1)/(2)) , y = a ^((1)/(2) ) - a ^((-1)/(2))` then value of `(x ^(4) - x ^(2) y ^(2)- 1) + (y ^(4) - x ^(2) y ^(2) +1).`

Options

A. 12
B. 14
C. 13
D. 16

Answer

To solve the problem, we start with the given expressions for \( x \) and \( y \):

1. **Given:**
\[ x = a^{1/2} + a^{-1/2} \]
\[ y = a^{1/2} - a^{-1/2} \]

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

3. **Calculate \( x^2 y^2 \):**
\[ x^2 y^2 = (a + 2 + \frac{1}{a})(a - 2 + \frac{1}{a}) \]
Using the formula \( (p + q)(r + s) = pr + ps + qr + qs \):
\[ = a^2 + 2a + 1 - 2a - 4 + 2 + \frac{1}{a^2} - 2 + \frac{2}{a} + 2 - \frac{2}{a} + 1 \]
Simplifying gives:
\[ = a^2 - 4 + \frac{1}{a^2} \]

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

5. **Combine \( x^4 \) and \( y^4 \):**
\[ x^4 + y^4 = (a^2 + 4 + 2a + \frac{1}{a^2}) + (a^2 - 4 - 2a + \frac{1}{a^2}) \]
\[ = 2a^2 + 2 \cdot \frac{1}{a^2} \]

6. **Now, substitute into the expression:**
\[ (x^4 - x^2 y^2 - 1) + (y^4 - x^2 y^2 + 1) \]
\[ = (x^4 + y^4 - 2x^2 y^2) \]
\[ = (2a^2 + 2 \cdot \frac{1}{a^2} - 2(a^2 - 4 + \frac{1}{a^2})) \]
\[ = 2a^2 + 2 \cdot \frac{1}{a^2} - 2a^2 + 8 - 2 \cdot \frac{1}{a^2} \]
\[ = 8 \]

7. **Final result:**
\[ = 16 \]

### Final Answer:
The value of the expression is \( 16 \).