If `x=9+4sqrt(5)` and `xy=1` , then find the value of `((1)/(x^(2))+(1)/(y^(2)))` .

Answer

To solve the problem step by step, we start with the given values and equations:

1. **Given Values**:
- \( x = 9 + 4\sqrt{5} \)
- \( xy = 1 \)

2. **Finding \( y \)**:
Since \( xy = 1 \), we can express \( y \) as:
\[ y = \frac{1}{x} = \frac{1}{9 + 4\sqrt{5}} \]

3. **Rationalizing \( y \)**:
To rationalize the denominator, we multiply the numerator and the denominator by the conjugate of the denominator:
\[ y = \frac{1}{9 + 4\sqrt{5}} \cdot \frac{9 - 4\sqrt{5}}{9 - 4\sqrt{5}} = \frac{9 - 4\sqrt{5}}{(9 + 4\sqrt{5})(9 - 4\sqrt{5})} \]

4. **Calculating the Denominator**:
Using the difference of squares:
\[ (9 + 4\sqrt{5})(9 - 4\sqrt{5}) = 9^2 - (4\sqrt{5})^2 = 81 - 80 = 1 \]
Thus, we have:
\[ y = 9 - 4\sqrt{5} \]

5. **Finding \( \frac{1}{x^2} \) and \( \frac{1}{y^2} \)**:
- First, calculate \( \frac{1}{x^2} \):
\[ \frac{1}{x^2} = \frac{1}{(9 + 4\sqrt{5})^2} \]
We will need to expand \( (9 + 4\sqrt{5})^2 \):
\[ (9 + 4\sqrt{5})^2 = 9^2 + 2 \cdot 9 \cdot 4\sqrt{5} + (4\sqrt{5})^2 = 81 + 72\sqrt{5} + 80 = 161 + 72\sqrt{5} \]
Therefore:
\[ \frac{1}{x^2} = \frac{1}{161 + 72\sqrt{5}} \]

- Now calculate \( \frac{1}{y^2} \):
\[ \frac{1}{y^2} = \frac{1}{(9 - 4\sqrt{5})^2} \]
Expanding \( (9 - 4\sqrt{5})^2 \):
\[ (9 - 4\sqrt{5})^2 = 9^2 - 2 \cdot 9 \cdot 4\sqrt{5} + (4\sqrt{5})^2 = 81 - 72\sqrt{5} + 80 = 161 - 72\sqrt{5} \]
Thus:
\[ \frac{1}{y^2} = \frac{1}{161 - 72\sqrt{5}} \]

6. **Finding \( \frac{1}{x^2} + \frac{1}{y^2} \)**:
Now we can add \( \frac{1}{x^2} \) and \( \frac{1}{y^2} \):
\[ \frac{1}{x^2} + \frac{1}{y^2} = \frac{1}{161 + 72\sqrt{5}} + \frac{1}{161 - 72\sqrt{5}} \]
To add these fractions, we find a common denominator:
\[ = \frac{(161 - 72\sqrt{5}) + (161 + 72\sqrt{5})}{(161 + 72\sqrt{5})(161 - 72\sqrt{5})} \]
The numerator simplifies to:
\[ 161 - 72\sqrt{5} + 161 + 72\sqrt{5} = 322 \]
The denominator, as calculated before, is:
\[ (161 + 72\sqrt{5})(161 - 72\sqrt{5}) = 161^2 - (72\sqrt{5})^2 = 25921 - 25920 = 1 \]
Thus:
\[ \frac{1}{x^2} + \frac{1}{y^2} = \frac{322}{1} = 322 \]

7. **Final Answer**:
The value of \( \frac{1}{x^2} + \frac{1}{y^2} \) is:
\[ \boxed{322} \]