The question in the image asks to simplify the expression $\frac{6}{3 - 2\sqrt{3}}$ and find its equivalent among the given options.

To solve this, we will use the method of rationalizing the denominator. The denominator $3 - 2\sqrt{3}$ can be rationalized by multiplying both the numerator and the denominator by its conjugate, $3 + 2\sqrt{3}$.

Let's simplify the expression step by step:

$\frac{6}{3 - 2\sqrt{3}} \times \frac{3 + 2\sqrt{3}}{3 + 2\sqrt{3}} = \frac{6(3 + 2\sqrt{3})}{(3 - 2\sqrt{3})(3 + 2\sqrt{3})}$

The denominator simplifies using the difference of squares formula:

$(3 - 2\sqrt{3})(3 + 2\sqrt{3}) = 3^2 - (2\sqrt{3})^2 = 9 - 4 \times 3 = 9 - 12 = -3$

Now, simplify the numerator:

$6(3 + 2\sqrt{3}) = 18 + 12\sqrt{3}$

So, the expression becomes:

$\frac{18 + 12\sqrt{3}}{-3} = \frac{18}{-3} + \frac{12\sqrt{3}}{-3} = -6 - 4\sqrt{3}$

The simplified form is $-6 - 4\sqrt{3}$, which corresponds to option b.

Would you like further details or explanations? Here are five related questions to consider:

1. How do you multiply by the conjugate to rationalize a denominator?
2. What is the difference of squares formula, and how is it used here?
3. How can you simplify radicals after rationalizing a denominator?
4. What are the key steps in recognizing when to use conjugates in simplifying fractions?
5. How would the solution change if the problem involved adding instead of subtracting within the square root?

Tip: Always double-check your final expression after rationalizing to ensure that the numerator and denominator have been simplified correctly.