If $a=\frac{2+\sqrt{3}}{2-\sqrt{3}}$ and $b = \frac{2-\sqrt{3}}{2+\sqrt{3}}$, then the value of $a^2 + b^2 +ab$ is:

195

$a=\frac{2+\sqrt{3}}{2-\sqrt{3}}$

$b = \frac{2-\sqrt{3}}{2+\sqrt{3}}$,

then the value of $a^2 + b^2 +ab$

When a fraction is in the form of $m=\frac{a+\sqrt{b}}{a-\sqrt{b}}$ and $m=\frac{a-\sqrt{b}}{a+\sqrt{b}}$ the difference between the square of a and square of b is equal to 1 then we can say, m = (a + b)²

and also n = (a - b)²

So, a = (2 + \(\sqrt {3}\))² = 7 + 2\(\sqrt {3}\)

b = 7 - 2\(\sqrt {3}\)

and ab = 1

So, $a^2 + b^2 +ab$  = (a + b)² - ab

= ( 7 + 2\(\sqrt {3}\) + 7 - 2\(\sqrt {3}\))² - 1 = 196 - 1 = 195