Answer
$\frac{x^2+2x+4}{x+2}$ and $x\ne2.$
Work Step by Step
The product of a binomial sum and a binomial difference is: $(A+B)(A-B)=A^2-B^2$.
The formula for factoring the sum of two cubes is: $A^3+B^3=(A+B)(A^2-AB+B^2)$.
The formula for factoring the difference of two cubes is: $A^3-B^3=(A-B)(A^2+AB+B^2)$.
Hence here: $\frac{x^3-8}{x^2-4}=\frac{x^3-2^3}{x^2-2^2}=\frac{(x-2)(x^2+2x+4)}{(x+2)(x-2)}=\frac{x^2+2x+4}{x+2}$ and $x\ne2.$
