The problem asks us to simplify the expression $\frac{3}{1 + \sqrt{3}}$. This requires rationalizing the denominator.

1. Problem Description

The problem asks us to simplify the expression $\frac{3}{1 + \sqrt{3}}$. This requires rationalizing the denominator.

2. Solution Steps

To rationalize the denominator, we multiply both the numerator and denominator by the conjugate of the denominator, which is $1 - \sqrt{3}$.

$\frac{3}{1 + \sqrt{3}} = \frac{3}{1 + \sqrt{3}} \cdot \frac{1 - \sqrt{3}}{1 - \sqrt{3}}$

$\frac{3}{1 + \sqrt{3}} \cdot \frac{1 - \sqrt{3}}{1 - \sqrt{3}} = \frac{3(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})}$

Using the difference of squares formula, $(a+b)(a-b) = a^2 - b^2$, we can simplify the denominator:

$(1 + \sqrt{3})(1 - \sqrt{3}) = 1^2 - (\sqrt{3})^2 = 1 - 3 = -2$

So, the expression becomes:

$\frac{3(1 - \sqrt{3})}{-2} = \frac{3 - 3\sqrt{3}}{-2} = \frac{-3 + 3\sqrt{3}}{2}$

Which can also be written as:

$\frac{3\sqrt{3} - 3}{2}$

3. Final Answer

$\frac{3\sqrt{3} - 3}{2}$