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The given expression to factor is:

\[
64 - p^6
\]

This is a **difference of squares** problem because \(64 = (8)^2\) and \(p^6 = (p^3)^2\). The general form for the difference of squares is:

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

For this expression, let \(a = 8\) and \(b = p^3\), so we can factor it as:

\[
64 - p^6 = (8 - p^3)(8 + p^3)
\]

Now, observe that the factor \(8 - p^3\) is a **difference of cubes**, which can be factored further. The general formula for the difference of cubes is:

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

Here, \(8 = 2^3\) and \(p^3\) is already in cubic form. So, we apply the formula with \(a = 2\) and \(b = p\):

\[
8 - p^3 = (2 - p)(4 + 2p + p^2)
\]

Therefore, the fully factored form of \(64 - p^6\) is:

\[
64 - p^6 = (2 - p)(4 + 2p + p^2)(8 + p^3)
\]

Would you like more details or have any questions? Here are five related problems:

1. Factor completely: \(x^6 - 64\).
2. Factor \(a^3 - 27\).
3. Factor \(x^4 - 81\).
4. Factor completely: \(125 - y^3\).
5. Simplify and factor: \(81 - x^4\).

**Tip:** Always look for patterns like the difference of squares or cubes to help factor large expressions efficiently.

Learn how to factor the algebraic expression 64 - p^6 using difference of squares and difference of cubes. This solution involves breaking down the problem into simpler forms and applying standard formulas. Ideal for high school students in Grades 9-12.

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