If `a=(2 + sqrt(3))/(2-sqrt(3)) and b = (2-sqrt(3))/(2+ sqrt(3))`, then the value of `a^(2)+b^(2)+ab` is :

Options

A. A)195
B. B)185
C. C)196
D. D)186

Answer

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

\[ a = \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \quad \text{and} \quad b = \frac{2 - \sqrt{3}}{2 + \sqrt{3}}. \]

### Step 1: Simplify \( a \) and \( b \)

First, we will simplify \( a \):

\[ a = \frac{2 + \sqrt{3}}{2 - \sqrt{3}}. \]

To simplify \( a \), we multiply the numerator and denominator by the conjugate of the denominator:

\[ a = \frac{(2 + \sqrt{3})(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})} = \frac{(2 + \sqrt{3})^2}{2^2 - (\sqrt{3})^2}. \]

Calculating the denominator:

\[ 2^2 - (\sqrt{3})^2 = 4 - 3 = 1. \]

Calculating the numerator:

\[ (2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}. \]

Thus,

\[ a = 7 + 4\sqrt{3}. \]

Now, we simplify \( b \):

\[ b = \frac{2 - \sqrt{3}}{2 + \sqrt{3}}. \]

Using the conjugate of the denominator again:

\[ b = \frac{(2 - \sqrt{3})(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{(2 - \sqrt{3})^2}{2^2 - (\sqrt{3})^2}. \]

The denominator remains the same:

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

Calculating the numerator:

\[ (2 - \sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3}. \]

Thus,

\[ b = 7 - 4\sqrt{3}. \]

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

Now, we calculate \( a + b \):

\[ a + b = (7 + 4\sqrt{3}) + (7 - 4\sqrt{3}) = 14. \]

Next, we calculate \( ab \):

\[ ab = \left( \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \right) \left( \frac{2 - \sqrt{3}}{2 + \sqrt{3}} \right) = \frac{(2 + \sqrt{3})(2 - \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})} = \frac{2^2 - (\sqrt{3})^2}{2^2 - (\sqrt{3})^2} = \frac{1}{1} = 1. \]

### Step 3: Calculate \( a^2 + b^2 + ab \)

Using the identity:

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

we can substitute the values we found:

\[ a^2 + b^2 = (14)^2 - 2(1) = 196 - 2 = 194. \]

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

\[ a^2 + b^2 + ab = 194 + 1 = 195. \]

### Final Answer

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

\[ \boxed{195}. \]