Answer
$\left( y-\dfrac{1}{10} \right) \left( y^2+\dfrac{1}{10}y+\dfrac{1}{100}\right)$
Work Step by Step
Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of 2 cubes, the factored form of the given expression, $
y^3-\dfrac{1}{1000}
,$ is
\begin{array}{l}
\left( y-\dfrac{1}{10} \right)\left[ (y)^2-(y)\left( -\dfrac{1}{10} \right)+\left( -\dfrac{1}{10} \right)^2\right]
\\\\=
\left( y-\dfrac{1}{10} \right) \left( y^2+\dfrac{1}{10}y+\dfrac{1}{100}\right)
.\end{array}
