If  $$\frac{4\sqrt{6} - 6\sqrt{2}}{\sqrt{96} + \sqrt{72}} \times \frac{1}{7 + \sqrt{48}} = a + b\sqrt{3}$$​, then what is the value of (a + b) ?

Correct option is C

Given:

​$$\frac{4\sqrt{6}-6\sqrt{2}}{\sqrt{96}+\sqrt{72}}\times\frac{1}{7+\sqrt{48}}=a+b\sqrt{3}$$

Solution:

Simplifying LHS;

​Now comparing with $$a+b\sqrt{3}$$​​

a = 97, b = -56

Thus, 

a + b = 97 - 56 = 41