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

Options

A. `sqrt(5)`
B. `2sqrt(5)`
C. `3sqrt(5)`
D. `4sqrt(5)`

Answer

To solve the problem, we need to find the value of \( x^2 - y^2 \) given:

\[ x = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} \]
\[ y = \frac{\sqrt{5} - 1}{\sqrt{5} + 1} \]

### Step 1: Use the difference of squares formula
We can use the identity \( a^2 - b^2 = (a - b)(a + b) \) to find \( x^2 - y^2 \):

\[ x^2 - y^2 = (x - y)(x + y) \]

### Step 2: Calculate \( x + y \)
First, we calculate \( x + y \):

\[ x + y = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} + \frac{\sqrt{5} - 1}{\sqrt{5} + 1} \]

To add these fractions, we need a common denominator, which is \( (\sqrt{5} - 1)(\sqrt{5} + 1) = 5 - 1 = 4 \):

\[ x + y = \frac{(\sqrt{5} + 1)^2 + (\sqrt{5} - 1)^2}{4} \]

Calculating the numerators:

\[ (\sqrt{5} + 1)^2 = 5 + 2\sqrt{5} + 1 = 6 + 2\sqrt{5} \]
\[ (\sqrt{5} - 1)^2 = 5 - 2\sqrt{5} + 1 = 6 - 2\sqrt{5} \]

Now, adding these:

\[ x + y = \frac{(6 + 2\sqrt{5}) + (6 - 2\sqrt{5})}{4} = \frac{12}{4} = 3 \]

### Step 3: Calculate \( x - y \)
Next, we calculate \( x - y \):

\[ x - y = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} - \frac{\sqrt{5} - 1}{\sqrt{5} + 1} \]

Using the same common denominator:

\[ x - y = \frac{(\sqrt{5} + 1)^2 - (\sqrt{5} - 1)^2}{4} \]

Calculating the numerator:

\[ (\sqrt{5} + 1)^2 - (\sqrt{5} - 1)^2 = (6 + 2\sqrt{5}) - (6 - 2\sqrt{5}) = 4\sqrt{5} \]

Thus,

\[ x - y = \frac{4\sqrt{5}}{4} = \sqrt{5} \]

### Step 4: Substitute back into the difference of squares formula
Now we can substitute back into the difference of squares formula:

\[ x^2 - y^2 = (x - y)(x + y) = (\sqrt{5})(3) = 3\sqrt{5} \]

### Final Answer
The value of \( x^2 - y^2 \) is:

\[ \boxed{3\sqrt{5}} \]