3B. Simplify the algebraic rational expression

$$
A=\frac{x^{3}+x^{2}+3 x^{2} y+2 x y+3 x y^{2}+y^{2}+y^{3}}{2 x^{3}+x^{2} y-4 x y^{2}-3 y^{2}}
$$
Solution. Let's factor the numerator:

$$
\begin{aligned}
x^{3}+x^{2}+3 x^{2} y+2 x y+ & 3 x y^{2}+y^{2}+y^{3}= \\
& =x^{3}+x^{2}+x^{2} y+2 x^{2} y+2 x y+2 x y^{2}+x y^{2}+y^{2}+y^{3} \\
& =x^{2}(x+1+y)+2 x y(x+1+y)+y^{2}(x+1+y) \\
& =(x+y)^{2}(x+1+y)
\end{aligned}
$$

Now let's try to factor the denominator so that one of the factors is \(x+y\):

$$
\begin{aligned}
2 x^{3}+x^{2} y-4 x y^{2}-3 y^{2} & =2 x^{3}+2 x^{2} y-x^{2} y-x^{2} y-3 x y^{2}-3 y^{2} \\
& =2 x^{2}(x+y)-x y(x+y)-3 y^{2}(x+y) \\
& =(x+y)\left(2 x^{2}-x y-3 y^{2}\right)
\end{aligned}
$$

We factor the polynomial \(2 x^{2}-x y-3 y^{2}\):

$$
2 x^{2}-x y-3 y^{2}=2 x^{2}+2 x y-3 x y-3 y^{2}=2 x(x+y)-3 y(x+y)=(x+y)(2 x-3 y)
$$

Finally, we get

$$
A=\frac{(x+y)^{2}(x+y+1)}{(x+y)^{2}(2 x-3 y)}=\frac{x+y+1}{2 x-3 y}
$$