If $$\frac{2 + \sqrt{3}}{2 - \sqrt{3}} + \frac{2 - \sqrt{3}}{2 + \sqrt{3}} + \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = a + b \sqrt{3}$$, then $$a + 4b =$$

Multiplying both numerator and denominator of first term by $$\left(2+\sqrt{\ 3}\right)$$, the term changes to $$\left(2+\sqrt{\ 3}\right)^2$$
Multiplying both numerator and denominator of second term by $$\left(2-\sqrt{\ 3}\right)$$, the term changes to $$\left(2-\sqrt{\ 3}\right)^2$$
Multiplying both numerator and denominator of third term by $$\left(\sqrt{\ 3}-1\right)$$, the term changes to $$\frac{\left(\sqrt{\ 3}-1\right)^2}{2}$$

Opening up the squares and adding up the terms, we get $$16-\sqrt{\ 3}$$, hence, a = 16, b = -1
Hence, a+4b becomes 12