Answer
$\color{blue}{y^2+3y+9}$
Work Step by Step
The given expression is equivalent to:
$=\dfrac{y^3-3^3}{y-3}$
Factor the numerator and the denominator then cancel the common factors. Factor the numerator using the formula $a^3+-b^3=(a-b)(a^2+ab+b^2)$ with $a=y$ and $b=3$.
:
$\require{cancel}
=\dfrac{(y-3)(y^2+3y+9)}{y-3}
\\=\dfrac{\cancel{(y-3)}(y^2+3y+9)}{\cancel{y-3}}
\\=\color{blue}{y^2+3y+9}$
