Question 148988
difference of cubes= a^3-b^3

a^3-b^3=(a-b)(a^2+ab+b^2)


in x^3-8, a=cube root of x^3 (or x) and b=cube root of 8 (or 2)
plug those in for a and b


So x^3-8/x-2 becomes
(x-2)(x^2+2x+4)/(x-2)


Now cancel the x-2 expressions and get
{{{(x^2+2x+4)}}}


This is the answer, but if you wish to factor it, you get:
{{{(x+2)^2-2x}}}

since {{{(x+2)^2=x^2+4x+4}}}
subtract 2x and get {{{x^2+2x+4}}}