98. If $$ x=\frac{4}{\sqrt{7}-\sqrt{3}} $$ and $$ y^{-1}=\frac{1}{\sqrt{7}-\sqrt{3}} $$ , prove that $$ x $$ and $$ y $$ are conjugate numbers, then find the value of : $$ x^{2} y^{2} $$

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To prove that $$ x $$ and $$ y $$ are conjugate numbers, we start with the given expressions for $$ x $$ and $$ y^{-1} $$.

1. **Calculate $$ x $$**:
   $$
   x = \frac{4}{\sqrt{7} - \sqrt{3}}
   $$
   To simplify $$ x $$, we can multiply the numerator and denominator by the conjugate of the denominator:
   $$
   x = \frac{4(\sqrt{7} + \sqrt{3})}{(\sqrt{7} - \sqrt{3})(\sqrt{7} + \sqrt{3})}
   $$
   The denominator simplifies as follows:
   $$
   (\sqrt{7} - \sqrt{3})(\sqrt{7} + \sqrt{3}) = 7 - 3 = 4
   $$
   Thus, we have:
   $$
   x = \frac{4(\sqrt{7} + \sqrt{3})}{4} = \sqrt{7} + \sqrt{3}
   $$

2. **Calculate $$ y $$**:
   Given $$ y^{-1} = \frac{1}{\sqrt{7} - \sqrt{3}} $$, we can find $$ y $$ by taking the reciprocal:
   $$
   y = \sqrt{7} - \sqrt{3}
   $$

3. **Check if $$ x $$ and $$ y $$ are conjugates**:
   We have:
   $$
   x = \sqrt{7} + \sqrt{3}, \quad y = \sqrt{7} - \sqrt{3}
   $$
   Since $$ x $$ and $$ y $$ are of the form $$ a + b $$ and $$ a - b $$ respectively, where $$ a = \sqrt{7} $$ and $$ b = \sqrt{3} $$, they are indeed conjugate numbers.

4. **Calculate $$ x^2 y^2 $$**:
   We can use the identity $$ (xy)^2 = x^2 y^2 $$. First, we find $$ xy $$:
   $$
   xy = (\sqrt{7} + \sqrt{3})(\sqrt{7} - \sqrt{3}) = 7 - 3 = 4
   $$
   Therefore:
   $$
   (xy)^2 = 4^2 = 16
   $$
   Thus, we have:
   $$
   x^2 y^2 = 16
   $$

5. **Final answer**:
   $$
   \boxed{16}
   $$
$$ x^2 y^2 = 16 $$

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, prove that and are conjugate numbers, then find the value of :

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