If `a=(sqrt(2)+1)/(sqrt(2)-1)"and"b=(sqrt(2)-1)/(sqrt(2)+1)` then value of `a^(2)+ab+b^(2)` is

Options

A. `70`
B. `35`
C. `40`
D. `34`

Answer

To solve the problem, we need to find the value of \( a^2 + ab + b^2 \) given that:

\[ a = \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \quad \text{and} \quad b = \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \]

### Step 1: Calculate \( a + b \)

First, we will find \( a + b \):

\[ a + b = \frac{\sqrt{2} + 1}{\sqrt{2} - 1} + \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \]

To add these fractions, we need a common denominator:

\[ a + b = \frac{(\sqrt{2} + 1)^2 + (\sqrt{2} - 1)^2}{(\sqrt{2} - 1)(\sqrt{2} + 1)} \]

Calculating the denominator:

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

Now, calculating the numerator:

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

Adding these:

\[ 3 + 2\sqrt{2} + 3 - 2\sqrt{2} = 6 \]

Thus,

\[ a + b = \frac{6}{1} = 6 \]

### Step 2: Calculate \( ab \)

Now, we will find \( ab \):

\[ ab = \left(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\right) \left(\frac{\sqrt{2} - 1}{\sqrt{2} + 1}\right) = \frac{(\sqrt{2} + 1)(\sqrt{2} - 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = 1 \]

### Step 3: Use the identity to find \( a^2 + b^2 \)

We know that:

\[ a^2 + b^2 = (a + b)^2 - 2ab \]

Substituting the values we found:

\[ a^2 + b^2 = 6^2 - 2 \cdot 1 = 36 - 2 = 34 \]

### Step 4: Calculate \( a^2 + ab + b^2 \)

Now we can find \( a^2 + ab + b^2 \):

\[ a^2 + ab + b^2 = a^2 + b^2 + ab = 34 + 1 = 35 \]

### Final Answer

Thus, the value of \( a^2 + ab + b^2 \) is:

\[ \boxed{35} \]