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)#
