To solve the expression $\frac{(2x^2 y^{-1})^7}{(4x^2)^3}$, we follow these steps:

Step 1: Simplify the numerator and denominator separately.

• Numerator: $(2x^2 y^{-1})^7$

  Apply the power to each factor inside the parentheses: $(2^7)(x^{2 \cdot 7})(y^{-1 \cdot 7}) = 128x^{14}y^{-7}$

• Denominator: $(4x^2)^3$

  Again, apply the power to each factor: $(4^3)(x^{2 \cdot 3}) = 64x^6$

Step 2: Combine the simplified numerator and denominator.

$\frac{128x^{14}y^{-7}}{64x^6}$

Step 3: Simplify the fraction.

• Simplify the coefficients: $\frac{128}{64} = 2$
• Simplify the powers of $x$: $\frac{x^{14}}{x^6} = x^{14-6} = x^8$

So, the expression becomes: $2x^8 y^{-7}$

Step 4: Express the answer in the simplest form.

• Since $y^{-7} = \frac{1}{y^7}$, the final answer is: $\frac{2x^8}{y^7}$

This is the simplified form of the given expression.

Would you like more details on any of these steps or have any questions?

Here are 5 related questions to expand your understanding:

1. How would the solution change if the exponent in the numerator were different, say $(2x^2 y^{-1})^5$?
2. Can you solve a similar problem with different variables, such as $\frac{(3a^3 b^{-2})^4}{(9a^2)^2}$?
3. What would be the effect of having a negative exponent in the denominator?
4. How would you simplify $\frac{(5x^3 y^2)^{4}}{(25x y^{-1})^2}$?
5. What are the steps to simplify expressions involving fractional exponents?

Tip: Always apply the exponent rules carefully, especially when dealing with negative exponents and coefficients, to avoid mistakes in simplification.