To factor the expression $x^2 y^2 - 81$ completely, we can identify it as a difference of squares. A difference of squares follows the form:

$a^2 - b^2 = (a - b)(a + b)$

Step-by-Step Solution

1. Rewrite the terms in square form: $x^2 y^2 = (xy)^2 \quad \text{and} \quad 81 = 9^2$

2. Now, rewrite the expression using these squares: $(xy)^2 - 9^2$

3. Apply the difference of squares formula: $(xy)^2 - 9^2 = (xy - 9)(xy + 9)$

Final Answer

The complete factorization of $x^2 y^2 - 81$ is: $(xy - 9)(xy + 9)$

Would you like further details on this solution or have any other questions?

Here are five related questions to explore further:

1. How do you recognize a difference of squares in other algebraic expressions?
2. Can a difference of squares formula be applied if the terms are not perfect squares?
3. What are some real-life applications of factoring differences of squares?
4. How would you factor a binomial that is a sum of squares, if possible?
5. How does factoring help simplify complex algebraic expressions?

Tip: Always check if an expression can be rewritten as a difference of squares, as this often simplifies the factorization process.