Answer
$\left(p+\dfrac{1}{5}\right)\left(p^2-\dfrac{p}{5}+\dfrac{1}{25}\right)$
Work Step by Step
Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of the sum of 2 cubes, then,
\begin{array}{l}
p^3+\dfrac{1}{25}
\\\\=
\left[(p)+\left(\dfrac{1}{5} \right) \right]\left[(p)^2-p\left(\dfrac{1}{5} \right)+\left(\dfrac{1}{5}\right)^2 \right]
\\\\=
\left(p+\dfrac{1}{5}\right)\left(p^2-\dfrac{p}{5}+\dfrac{1}{25}\right)
.\end{array}
