Answer
$\frac{(y+4)(y+2)}{(y-6)(y^2+4y+16)}$.
Work Step by Step
The given expression is
$=\frac{y^2-16}{y^3-64}\div \frac{y^2-3y-18}{y^2+5y+6}$
Factor $y^2-16$.
$=y^2-4^2$
Use the algebraic identity $a^2-b^2=(a+b)(a-b)$.
$(y+4)(y-4)$
Factor $y^3-64$.
$=y^3-4^3$
Use the algebraic identity $a^3-b^3=(a-b)(a^2+ab+b^2)$.
$(y-4)(y^2+4y+16)$
Factor $y^2-3y-18$.
Rewrite the middle term $-3y$ as $-6y+3y$.
$=y^2-6y+3y-18$
Group terms.
$=(y^2-6y)+(3y-18)$
Factor each term.
$=y(y-6)+3(y-6)$
Factor out $(y-6)$.
$=(y-6)(y+3)$
.
Factor $y^2+5y+6$.
Rewrite the middle term $5y$ as $3y+2y$.
$=y^2+3y+2y+6$
Group terms.
$=(y^2+3y)+(2y+6)$
Factor each term.
$=y(y+3)+2(y+3)$
Factor out $(y+3)$.
$=(y+3)(y+2)$
Substitute the factor into the given expression.
$=\frac{(y+4)(y-4)}{(y-4)(y^2+4y+16)}\div \frac{(y-6)(y+3)}{(y+3)(y+2)}$
$=\frac{(y+4)(y-4)}{(y-4)(y^2+4y+16)}\times \frac{(y+3)(y+2)}{(y-6)(y+3)}$
Cancel common terms.
$=\frac{(y+4)(y+2)}{(y-6)(y^2+4y+16)}$.
