If a = $$\frac{3+\sqrt5}{3-\sqrt5}$$​ and b = $$\frac{3-\sqrt5}{3+\sqrt5}$$, then find the value of $$\text{a}^2+\text{b}^2-\text{ab}$$​.

Correct option is A

Given:

a = $$\dfrac{3 + \sqrt{5}}{3 - \sqrt{5}}, \quad b = \dfrac{3 - \sqrt{5}}{3 + \sqrt{5}}$$​​
Find the value of $$a^2 + b^2 - ab.$$​​

Formula Used:

​$$(a + b)^2 = a^2 + b^2 + 2ab$$​​

​$$(a - b)^2 = a^2 + b^2 - 2ab$$​​

Solution:

Compute a + b

a + b =$$\dfrac{3 + \sqrt{5}}{3 - \sqrt{5}} + \dfrac{3 - \sqrt{5}}{3 + \sqrt{5}}$$​

=$$\dfrac{(3 + \sqrt{5})^2 + (3 - \sqrt{5})^2}{(3 - \sqrt{5})(3 + \sqrt{5})}$$​

=$$\frac { 9 + 5 + 6 \sqrt 5 + 9 + 5 - 6 \sqrt 5 }{ 9-5}$$

=$$\frac {28}4$$ =7​

Compute ab

ab =$$\left( \dfrac{3 + \sqrt{5}}{3 - \sqrt{5}} \right) \left( \dfrac{3 - \sqrt{5}}{3 + \sqrt{5}} \right) = 1$$​

Compute $$a^2 + b^2 - ab$$​​

​$$a^2 + b^2 = (a + b)^2 - 2ab = 7^2 - 2(1) = 49 - 2 = 47$$​​

Now ,

$$a^2 + b^2 - ab = 47 - 1 = 46$$​