Answer
$y^2+2y+4$
Work Step by Step
$y^3-8$ can be written as $y^3-2^3$, which is a difference of two cubes.
RECALL:
A sum or difference of two cubes can be factored using either of the following formulas:
(1) $a^3+b^3=(a-b)(a^2-ab+b^2)$
(2) $a^3-b^3=(a-b)(a^2+ab+b^2)$
Using formula (2) above with $a=y$ and $b=2$ gives:
$\dfrac{y^3-8}{y-2}
\\= \dfrac{(y-2)(y^2+y(2)+2^2)}{y-2}
\\=\dfrac{(y-2)(y^2+2y+4)}{y-2}$
Cancel out the common factor to obtain:
$\require{cancel}
\\=\dfrac{\cancel{(y-2)}(y^2+2y+4)}{\cancel{y-2}}
\\=y^2+2y+4$
