Let's solve the expression shown in the uploaded image step by step.

The given expression is:

$\frac{3^6 \times (24)^{-2}}{18^3 \times (2^{-3})^4}$

Step 1: Simplify powers

We will simplify the terms involving exponents first.

• Simplify $(2^{-3})^4$: $(2^{-3})^4 = 2^{-12}$

• Express 18 in terms of its prime factors: $18 = 2 \times 3^2 \quad \Rightarrow \quad 18^3 = (2 \times 3^2)^3 = 2^3 \times 3^6$

Now the expression becomes:

$\frac{3^6 \times 24^{-2}}{2^3 \times 3^6 \times 2^{-12}}$

Step 2: Express 24 in terms of prime factors

$24 = 2^3 \times 3 \quad \Rightarrow \quad 24^{-2} = (2^3 \times 3)^{-2} = 2^{-6} \times 3^{-2}$

Now the expression becomes:

$\frac{3^6 \times 2^{-6} \times 3^{-2}}{2^3 \times 3^6 \times 2^{-12}}$

Step 3: Simplify the terms involving the same bases

• Simplify the powers of 2: $\frac{2^{-6}}{2^3 \times 2^{-12}} = 2^{-6 - 3 + 12} = 2^3$

• Simplify the powers of 3: $\frac{3^6 \times 3^{-2}}{3^6} = 3^{6 - 2 - 6} = 3^{-2}$

Now the expression becomes:

$2^3 \times 3^{-2}$

Step 4: Evaluate the result

$2^3 = 8 \quad \text{and} \quad 3^{-2} = \frac{1}{9}$

Thus:

$2^3 \times 3^{-2} = 8 \times \frac{1}{9} = \frac{8}{9}$

Final Answer

$\boxed{\frac{8}{9}}$

Would you like any further steps or clarifications? 😊

Here are 5 follow-up questions for deeper exploration:

1. How do the rules of negative exponents work in general?
2. How do you factor numbers into their prime factors efficiently?
3. What is the significance of the order of operations when simplifying powers?
4. Can you generalize how to simplify expressions with multiple exponents?
5. How would the result change if one of the exponents was zero?

Tip: When dealing with powers, always group similar bases to simplify calculations easily.