If ` x = ( sqrt(2) + 1)/(sqrt(2) - 1), y = ( sqrt(2) - 1)/( sqrt(2) + 1)` then the value of ` ( x^(2) + 6xy + y^(2))/(x^(2) - 6xy + y^(2)) ` is

Options

A. `(5)/( 7)`
B. `(5)/(3)`
C. `(10)/(7)`
D. `(10)/(3)`

Answer

To solve the problem, we need to find the value of

\[ \frac{x^2 + 6xy + y^2}{x^2 - 6xy + y^2} \]

given:

\[ x = \frac{\sqrt{2} + 1}{\sqrt{2} - 1}, \quad y = \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \]

### Step 1: Simplify \(x\) and \(y\)

**For \(x\):**

Multiply the numerator and denominator of \(x\) by \((\sqrt{2} + 1)\):

\[ x = \frac{(\sqrt{2} + 1)(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{(\sqrt{2} + 1)^2}{(\sqrt{2})^2 - 1^2} = \frac{2 + 2\sqrt{2} + 1}{2 - 1} = 3 + 2\sqrt{2} \]

**For \(y\):**

Multiply the numerator and denominator of \(y\) by \((\sqrt{2} - 1)\):

\[ y = \frac{(\sqrt{2} - 1)(\sqrt{2} - 1)}{(\sqrt{2} + 1)(\sqrt{2} - 1)} = \frac{(\sqrt{2} - 1)^2}{(\sqrt{2})^2 - 1^2} = \frac{2 - 2\sqrt{2} + 1}{2 - 1} = 3 - 2\sqrt{2} \]

### Step 2: Calculate \(x + y\) and \(xy\)

**Sum \(x + y\):**

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

**Product \(xy\):**

Using the difference of squares:

\[ xy = (3 + 2\sqrt{2})(3 - 2\sqrt{2}) = 3^2 - (2\sqrt{2})^2 = 9 - 8 = 1 \]

### Step 3: Substitute \(x + y\) and \(xy\) into the expression

We know:

\[ x^2 + y^2 = (x + y)^2 - 2xy = 6^2 - 2 \cdot 1 = 36 - 2 = 34 \]

Now we can substitute into the expression:

\[ x^2 + 6xy + y^2 = x^2 + y^2 + 6xy = 34 + 6 \cdot 1 = 34 + 6 = 40 \]

\[ x^2 - 6xy + y^2 = x^2 + y^2 - 6xy = 34 - 6 \cdot 1 = 34 - 6 = 28 \]

### Step 4: Calculate the final value

Now we compute:

\[ \frac{x^2 + 6xy + y^2}{x^2 - 6xy + y^2} = \frac{40}{28} = \frac{10}{7} \]

### Final Answer:

The value of

\[ \frac{x^2 + 6xy + y^2}{x^2 - 6xy + y^2} = \frac{10}{7} \]

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