#### Answer

$-y(3x^2+3xy+y^2)$

#### Work Step by Step

The expressions $
x^3
$ and $
(x+y)^3
$ are both perfect cubes (the cube root is exact). Hence, $
x^3-(x+y)^3
$ is a $\text{
difference
}$ of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(x)^3-(x+y)^3
\\\\=
[x-(x+y)][(x)^2+x(x+y)+(x+y)^2]
\\\\=
[x-x-y][x^2+x^2+xy+(x^2+2xy+y^2)]
\\\\=
-y(x^2+x^2+xy+x^2+2xy+y^2)
\\\\=
-y(3x^2+3xy+y^2)
.\end{array}