What is the limit of ##( x^3 – 8 )/ (x-2)## as x approaches 2?

Notice how you have a difference of two cubes.
##(x-y)(x^2 + xy + y^2) = x^3 + x^2y + xy^2 – x^2y – xy^2 – y^3 = x^3 – y^3##
##((x)^3 – (2)^3)/(x-2) = ((x-2)(x^2 + 2x + 4))/(x-2) = x^2 + 2x + 4##
Plug in ##2##:
##lim_(x->2)(x^3 – 8)/(x-2)= 4 + 4 + 4 = 12##