Let's go through the solution to this expression.

Given Expression:

$\frac{6^0}{(3^{-2} x^{-3} y^2)^2}$

Step 1: Simplify $6^0$

By the zero exponent rule, any non-zero number raised to the power of zero is 1: $6^0 = 1$

So, the expression now becomes: $\frac{1}{(3^{-2} x^{-3} y^2)^2}$

Step 2: Simplify the Denominator by Applying the Power of a Power Rule

Now, we need to square each term inside the parentheses: $(3^{-2} x^{-3} y^2)^2 = 3^{-2 \cdot 2} \cdot x^{-3 \cdot 2} \cdot y^{2 \cdot 2}$

This gives us: $= 3^{-4} \cdot x^{-6} \cdot y^4$

Step 3: Rewrite with Positive Exponents

To make all exponents positive, we move terms with negative exponents to the numerator: $= \frac{1}{\frac{1}{3^4 x^6 y^4}} = 3^4 x^6 y^4$

Step 4: Calculate $3^4$

Since $3^4 = 81$, the expression becomes: $81 x^6 y^4$

Final Answer:

$81 x^6 y^4$

If you have any questions about the steps, please let me know!

Follow-up Questions

1. How does the zero exponent rule work, and why does $6^0 = 1$?
2. What is the power of a power rule, and how is it applied here?
3. How would the answer change if the exponents were positive initially?
4. Why do we move terms with negative exponents to the numerator or denominator?
5. What would happen if we had another constant in the numerator, such as $7^0$?

Tip:

Always simplify terms with zero exponents first, as they often reduce the complexity of an expression significantly.