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##