Find a special factoring pattern for a difference of two sextics: (a^6-b^6).

first step:

(a^3 - b^3)(a^3 + b^3)

second step: we can factor the sum and difference of cubes, should be in your notes or text

= (a-b)(a^2 + ab + b^2)(a+b)(a^2 - ab + b^2)

To factor the difference of two sextics, (a^6 - b^6), we can use the pattern called "difference of cubes." This pattern is based on the identity: a^3 - b^3 = (a - b)(a^2 + ab + b^2).

To apply this pattern to a^6 - b^6, we need to rewrite it as a difference of cubes. We can rewrite a^6 as (a^3)^2 and b^6 as (b^3)^2. So, we have:

a^6 - b^6 = (a^3)^2 - (b^3)^2

Now, we can apply the difference of cubes pattern by setting a^3 = a and b^3 = b:

= (a^3 - b^3)(a^3 + b^3)

Substituting back a and b:

= (a - b)(a^3 + b^3)

Notice that a^3 + b^3 cannot be further factored using this pattern, so our final factored form is:

a^6 - b^6 = (a - b)(a^3 + b^3)

Therefore, the special factoring pattern for the difference of two sextics is (a - b)(a^3 + b^3).