Answer
$5\left(xy^2-\dfrac{1}{2}\right)\left(x^2y^4+\dfrac{xy^2}{2}+\dfrac{1}{4}\right)$
Work Step by Step
Factoring the $GCF=
5
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
5x^3y^6-\dfrac{5}{8}
\\\\=
5\left(x^3y^6-\dfrac{1}{8}\right)
.\end{array}
The expressions $
x^3y^6
$ and $
\dfrac{1}{8}
$ are both perfect cubes (the cube root is exact). Hence, $
x^3y^6-\dfrac{1}{8}
$ 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}
5\left[\left(xy^2\right)^3-\left(\dfrac{1}{2}\right)^3\right]
\\\\=
5\left(xy^2-\dfrac{1}{2}\right)\left[(xy^2)^2+xy^2\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2 \right]
\\\\=
5\left(xy^2-\dfrac{1}{2}\right)\left(x^2y^4+\dfrac{xy^2}{2}+\dfrac{1}{4}\right)
.\end{array}
