To solve the problem in the image, let's break down each part:

The question is asking for the value of: $(27 \times 2^{-3})^{\frac{1}{3}} \div \left(\frac{1}{2} \times 3\right)^{-2}.$

Let's solve it step-by-step.

Step 1: Simplify $(27 \times 2^{-3})^{\frac{1}{3}}$

1. Start by simplifying $27 \times 2^{-3}$:

   • $27 = 3^3$, so $27 \times 2^{-3} = 3^3 \times 2^{-3} = (3 \times 2^{-1})^3$.
   • Since $2^{-1} = \frac{1}{2}$, this becomes $\left(3 \times \frac{1}{2}\right)^3 = \left(\frac{3}{2}\right)^3$.

2. Now, raise this expression to the power of $\frac{1}{3}$:

   • $\left(\frac{3}{2}\right)^3$ raised to $\frac{1}{3}$ is $\frac{3}{2}$ because $3 \times \frac{1}{3} = 1$.

So, $(27 \times 2^{-3})^{\frac{1}{3}} = \frac{3}{2}$.

Step 2: Simplify $\left(\frac{1}{2} \times 3\right)^{-2}$

1. Simplify inside the parentheses: $\frac{1}{2} \times 3 = \frac{3}{2}$.
2. Now raise $\frac{3}{2}$ to the power of $-2$:
   • $\left(\frac{3}{2}\right)^{-2} = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$.

So, $\left(\frac{1}{2} \times 3\right)^{-2} = \frac{4}{9}$.

Step 3: Divide the results

Now, we need to calculate: $\frac{3}{2} \div \frac{4}{9}.$ To divide fractions, multiply by the reciprocal: $\frac{3}{2} \times \frac{9}{4} = \frac{3 \times 9}{2 \times 4} = \frac{27}{8}.$

Final Answer

The correct answer is: $\boxed{\frac{27}{8}}.$

So, the answer is option D.

Would you like more details on any part of this solution?

Here are 5 related questions you might be interested in:

1. What is the general process for simplifying expressions with negative exponents?
2. How do we simplify expressions with fractional exponents?
3. Why do we multiply by the reciprocal when dividing fractions?
4. What are some common mistakes to watch for with exponent rules?
5. How can we apply these techniques to more complex algebraic expressions?

Tip: When working with fractional and negative exponents, rewrite each component in a simpler form first. This often makes it easier to handle the entire expression.