To factor $x^6 - 64$, we can recognize that this is a difference of squares. It takes the form $a^2 - b^2$, which factors as $(a - b)(a + b)$.

In this case: $x^6 - 64 = (x^3)^2 - 8^2$ Now, apply the difference of squares formula: $(x^3 - 8)(x^3 + 8)$

Next, notice that both $x^3 - 8$ and $x^3 + 8$ are cubes. Specifically, these are differences and sums of cubes, which can be factored further:

1. $x^3 - 8$ is a difference of cubes, which factors as: $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$

2. $x^3 + 8$ is a sum of cubes, which factors as: $x^3 + 8 = (x + 2)(x^2 - 2x + 4)$

Thus, the full factorization of $x^6 - 64$ is: $x^6 - 64 = (x - 2)(x^2 + 2x + 4)(x + 2)(x^2 - 2x + 4)$

Summary:

The fully factored form of $x^6 - 64$ is: $(x - 2)(x + 2)(x^2 + 2x + 4)(x^2 - 2x + 4)$

Would you like further clarification or details on any step?

Here are some additional related questions:

1. How do you factor a difference of cubes?
2. What is the general formula for factoring a sum or difference of cubes?
3. How do you factor $x^3 - 27$ or $x^3 + 27$?
4. What is the significance of the expression $x^2 + 2x + 4$?
5. How do you recognize and factor higher-degree polynomials like $x^6 - a^6$?

Tip: When dealing with higher-degree polynomials like $x^6 - 64$, it's helpful to first identify if the expression can be rewritten as a difference or sum of squares or cubes, as these can often be factored further.