## Intermediate Algebra (6th Edition)

$y^2+2y+4$
$y^3-8$ 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: $\require{cancel} \\=\dfrac{\cancel{(y-2)}(y^2+2y+4)}{\cancel{y-2}} \\=y^2+2y+4$