factor completely: z^6-z^2

Take out the z^2

z^2 (z^4-1)
now the last term is a difference of two squares, so factor it.
z^2(z^2+1)(z^2-1)
again, the last term is a difference of squares again, so factor it also.

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To factor the expression z^6 - z^2 completely, we can look for common factors and apply some algebraic techniques. Let's break down the process step-by-step:

Step 1: Look for common factors.
In this case, we can see that z^2 is a common factor in both terms: z^6 and z^2. Let's factor it out:

z^6 - z^2 = z^2(z^4 - 1)

Step 2: Recognize a difference of squares.
Now, we can observe that z^4 - 1 is a difference of squares. We can write it as (z^2)^2 - 1^2.

Using the difference of squares formula, a^2 - b^2 = (a + b)(a - b), we can factorize (z^2)^2 - 1^2 as follows:
(z^2 + 1)(z^2 - 1)

Step 3: Factorize further.
At this point, we have factored z^6 - z^2 as z^2(z^2 + 1)(z^2 - 1). However, we can still factor z^2 - 1 using the difference of squares formula again.

Applying the difference of squares formula to z^2 - 1, we get:
(z + 1)(z - 1)

Therefore, the fully factored form of z^6 - z^2 is:
z^2(z^2 + 1)(z + 1)(z - 1)