Question 8167
Ok,this gets a little messy, so please bear with me.

Factor: {{{x^12 - 1}}} Start with the difference of two squares.

{{{x^12 - 1 = (x^6 - 1)(x^6 + 1)}}} NB: {{{x^12 = (x^6)^2}}}

Take the first factor: {{{(x^6 - 1)}}} and note that it is also the difference of two squares.

{{{(x^6 - 1) = (x^3 - 1)(x^3 + 1)}}}

Notice here that the 1st factor {{{(x^3 - 1)}}} is the difference of two cubes and that the second factor {{{(x^3 + 1)}}} is the sum of two cubes.  These can be factored using the formulas:

{{{(A^3 - B^3) = (A - B)(A^2 + AB + B^2)}}} and;
{{{(A^3 + B^3) = (A + B)(A^2 - AB + B^2)}}}

So, we have: 

{{{(x^3 - 1) = (x - 1)(x^2 + x + 1)}}} and;
{{{(x^3 + 1) = (x + 1)(x^2 - x + 1)}}}

Now, going back to the first line, let's look at the 2nd factor: {{{(x^6 + 1)}}} and factor this:

{{{(x^6 + 1) = (x^2+ 1)(x^4 - x^2 + 1)}}} 

Now we can put it all together:

{{{(x^12 - 1) = (x - 1)(x + 1)(x^2 + 1)(x^2 + x + 1)(x^2 - x + 1)(x^4 - x^2 + 1)}}} Whew!

You can verify this by multiplying all of these factors to see that you get {{{x^12 - 1)}}} I'll leave this as an exercise for the student.

A final note:

I have left the factor {{{(x^2 + 1)}}} in its unfactored form, not because it can't be factored, but because I don't know whether or not you have covered complex numbers yet. But, in case you have, here's how you would factor {{{x^2 + 1}}}

{{{(x^2 + 1) = (x - i)(x + i)}}} where: {{{i = sqrt(-1)}}} 

Using FOIL to check:

{{{(x - i)(x + i) = (x^2 + xi - xi - i^2)}}} NB: {{{i^2 = (sqrt(-1))^2}}} = -1

{{{(x - i)(x + i) = x^2 - (-1)}}} = {{{x^2 + 1)}}}