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).