Factor completely.
x, squared, y, to the power 6 , minus, 1
x 
2
 y 
6
 −1

The given expression is:

\[
x^2 y^6 - 1
\]

This is a difference of squares, since \( 1 = 1^2 \), and it can be factored using the formula for the difference of squares:

\[
a^2 - b^2 = (a - b)(a + b)
\]

In this case, \( a = xy^3 \) and \( b = 1 \), because:

\[
(x^2 y^6 - 1) = (xy^3)^2 - 1^2
\]

Now apply the difference of squares formula:

\[
(x^2 y^6 - 1) = (xy^3 - 1)(xy^3 + 1)
\]

So the completely factored form of \( x^2 y^6 - 1 \) is:

\[
(xy^3 - 1)(xy^3 + 1)
\]

Would you like further details on factoring or any other questions?

### Related Questions:
1. How do you factor higher-degree polynomials?
2. What are some common techniques for factoring differences of squares?
3. How does factoring help in solving polynomial equations?
4. Can the difference of cubes be factored similarly to the difference of squares?
5. How do you verify that a factorization is correct?

### Tip:
Always check if the expression is a difference of squares before trying other factoring techniques—it often simplifies the process.

Explore how to factor the expression x^2 y^6 − 1 using the difference of squares formula. This step-by-step guide covers key algebraic concepts like factoring and the difference of squares theorem, perfect for students in Grades 9-11.

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