#### Answer

$5(x+3y^2)(x^2-y)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Factor the $GCF$ of the terms. Then, group the terms of the given expression, $
5x^3+15x^2y^2-5xy-15y^3
,$ such that the factored form of the groupings will result to a factor common to the entire expression. Then, factor the $GCF$ in each group. Finally, factor the $GCF$ of the entire expression.
$\bf{\text{Solution Details:}}$
Using the $GCF=
5
,$, the expression above is equivalent to
\begin{array}{l}\require{cancel}
5\left( \dfrac{5x^3}{5}+\dfrac{15x^2y^2}{5}-\dfrac{5xy}{5}-\dfrac{15y^3}{5} \right)
\\\\=
5\left( x^3+3x^2y^2-xy-3y^3 \right)
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
5[(x^3+3x^2y^2)-(xy+3y^3)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
5[x^2(x+3y^2)-y(x+3y^2)]
.\end{array}
Factoring the $GCF=
(x+3y^2)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
5[(x+3y^2)(x^2-y)]
\\\\=
5(x+3y^2)(x^2-y)
.\end{array}