To solve the expression

$2^3 \times \left(\frac{3^3 \times 2^{-2}}{3^{-2} \times 2^4}\right)^3,$

we can follow these steps:

Step 1: Simplify Inside the Parentheses

First, look at the expression inside the parentheses:

$\frac{3^3 \times 2^{-2}}{3^{-2} \times 2^4}.$

Simplify the numerator:

• $3^3 = 27$
• $2^{-2} = \frac{1}{2^2} = \frac{1}{4}$

Therefore, the numerator becomes $27 \times \frac{1}{4} = \frac{27}{4}$.

Simplify the denominator:

• $3^{-2} = \frac{1}{3^2} = \frac{1}{9}$
• $2^4 = 16$

Therefore, the denominator becomes $\frac{1}{9} \times 16 = \frac{16}{9}$.

Now, substitute back:

$\frac{3^3 \times 2^{-2}}{3^{-2} \times 2^4} = \frac{\frac{27}{4}}{\frac{16}{9}}.$

Step 2: Divide the Fractions

$\frac{\frac{27}{4}}{\frac{16}{9}} = \frac{27}{4} \times \frac{9}{16} = \frac{27 \times 9}{4 \times 16} = \frac{243}{64}.$

Step 3: Apply the Cube to the Fraction

Now, cube the result from inside the parentheses:

$\left(\frac{243}{64}\right)^3 = \frac{243^3}{64^3}.$

Calculating the powers:

• $243^3 = (3^5)^3 = 3^{15}$
• $64^3 = (2^6)^3 = 2^{18}$

So,

$\left(\frac{243}{64}\right)^3 = \frac{3^{15}}{2^{18}}.$

Step 4: Multiply by $2^3$

Now, go back to the original expression:

$2^3 \times \frac{3^{15}}{2^{18}}.$

$= \frac{2^3 \times 3^{15}}{2^{18}}.$

Step 5: Combine Exponents

$= \frac{3^{15}}{2^{18-3}} = \frac{3^{15}}{2^{15}}.$

Final Answer

$\frac{3^{15}}{2^{15}} = \left(\frac{3}{2}\right)^{15}.$

Would you like more details or have any questions?

Here are five related questions to consider:

1. How would you simplify $\left(\frac{4^2 \times 5^{-1}}{2^{-3} \times 5^3}\right)^2$?
2. What is the result of $\frac{6^3 \times 2^{-4}}{3^{-3} \times 2^2}$?
3. How do you deal with negative exponents in complex fractions?
4. Can you express $(a^3 \times b^{-2})^2 \times (a^{-1} \times b^4)^{-1}$ in a simplified form?
5. How would the approach change if the original problem had a base of zero for any of the exponents?

Tip: Remember, when working with exponents, handling the negative exponents first can simplify the process, making multiplication and division more straightforward.