Factor x⁶ - y⁶ as difference of cubes

To factor \( x^6 - y^6 \) as a difference of cubes, we can start by recognizing that both \( x^6 \) and \( y^6 \) can be written as cubes:

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

So, we can rewrite the original expression as:

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

This is now in the form of a difference of cubes, \( a^3 - b^3 \), which factors as:

\[
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
\]

In our case, \( a = x^2 \) and \( b = y^2 \). Therefore, we can factor \( x^6 - y^6 \) as:

\[
x^6 - y^6 = (x^2 - y^2)\left((x^2)^2 + (x^2)(y^2) + (y^2)^2\right)
\]

Now, simplify the expression inside the parentheses:

\[
(x^2)^2 + (x^2)(y^2) + (y^2)^2 = x^4 + x^2y^2 + y^4
\]

So the factored form is:

\[
x^6 - y^6 = (x^2 - y^2)(x^4 + x^2y^2 + y^4)
\]

Finally, notice that \( x^2 - y^2 \) is a difference of squares, which can be further factored:

\[
x^2 - y^2 = (x - y)(x + y)
\]

So, the complete factorization of \( x^6 - y^6 \) is:

\[
x^6 - y^6 = (x - y)(x + y)(x^4 + x^2y^2 + y^4)
\]

Would you like any additional details or have any questions?

Here are 5 related questions:

1. How do you factor the sum of cubes?
2. What is the difference of squares, and how is it factored?
3. Can you factor \( x^8 - y^8 \) in a similar way?
4. How do you factor a trinomial such as \( x^4 + x^2y^2 + y^4 \)?
5. What is the general formula for factoring \( a^n - b^n \)?

**Tip:** Always look for common patterns such as the difference of squares or cubes when factoring complex polynomials.

Explore the step-by-step process of factoring the expression x^6 - y^6 by applying the difference of cubes and difference of squares formulas. This algebraic problem is ideal for students in Grades 9-11.

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