It could be both, actually.

You can use the properties of exponential powers to write those terms both as a difference of squares, and as a difference of cubes.

Since ##(a^x)^y = a^(xy)##, you can say that

##x^(12) = x^(6 * color(red)(2)) = (x^(6))^(color(red)(2))##

and

##y^(12) = (y^(6))^(color(red)(2)##

This means that you get

##x^(12) – y^(12) = (x^(6))^(2) – (y^(6))^(2) = (x^(6) – y^(6))(x^(6) + y^(6))##

Likewise,

##x^(12) = x^(4 * color(red)(3)) = (x^(4))^(color(red)(3))## and ##y^(12) = (y^(4))^(color(red)(3))##

So you can write

##x^(12) – y^(12) = (x^(4))^(3) – (y^(4))^(3) = (x^4 – y^4)[(x^(4))^2 + x^(4)y^(4) + (y^4)^(2)]##

##x^12 – y^12 = (x^4 – y^4)[x^8 + x^(4)y^4 + y^8]##

As you can see, you can simplify these expressions further. Here’s how you would factor this expression completely

##x^(12) – y^(12) = underbrace((x^6 – y^6))_(color(green)(“difference of two squares”)) * underbrace((x^6 + y^6))_(color(blue)(“sum of two cubes”)) = ##

##=underbrace((x^3 – y^3))_(color(green)(“difference of two cubes”)) * underbrace((x^3 + y^3))_(color(blue)(“sum of two cubes”)) * (x^2 + y^2)(x^4 + x^2 * y^2 + y^4) = ##

##=(x + y)(x^2 -xy + y^2) * (x-y)(x^2 + xy + y^2) * (x^2 + y^2)(x^4 + x^2 * y^2 + y^4)##

##x^12 – y^12 = (x + y)(x-y)(x^2 + y^2)(x^2 – xy + y^2)(x^2 + xy + y^2)(x^4 + x^2 y^2 + y^2)##