Answer
$y^2+3y+9$
$y\ne3$
Work Step by Step
Given.
$\frac{y^3-27}{y-3}$
Write 27 as $3^3$.
$=\frac{y^3-3^3}{y-3}$
Factor the sum of two cubes using the formula $a^3-b^3=(a-b)(a+ab+b^2)$.
$=\frac{(y-3)(y^2+3y+9)}{y-3}$
Cancel the common factor $y-3$.
$y^2+3y+9,y\ne3$
