Question 723008
The exponents are understood.
y^3 is  the usual way to type {{{y^3}}} and is y^2 the usual way to type {{{y^2}}}
What is not clear is what to do with that rational expression.
We can factor or divide. Maybe one of those two choices is what you wanted.
 
FACTORING:
{{{(y^3-y^2-3y-9)/(y+1)=(y-3)(y^2+2y+3)/(y+1)}}}
and we see that it cannot be simplified.
The possible rational zeros of {{{y^3-y^2-3y-9}}} are -9, -3, -1, 1, 3, and 9.
It was easy to try -3, -1, 1, and 3.
Since 3 was a zero {{{y-3}}} had to be a factor.
Dividing by {{{(y-3)}}} we get {{{y^2+2y+3}}} and
{{{y^2+2y+3}}} cannot be factored further because it has no zeros.
(There are no real solutions to {{{y^2+2y+3=0}}} ).
 
DIVIDING:
{{{(y^3-y^2-3y-9)/(y+1)=y^2-2y-1-8/(y+1)}}}
We get {{{y^2-2y-1}}} for a quotient and a remainder of {{{-8}}} .