$8 p^{3}+\frac{12}{5} p^{2}+\frac{6}{25} p+\frac{1}{125}$

$=(2 p)^{3}+3 \times(2 p)^{2} \times \frac{1}{5}+3 \times(2 p)+\left(\frac{1}{5}\right)^{2}+\left(\frac{1}{5}\right)^{3}$

$=(2 p)^{3}+\left(\frac{1}{5}\right)^{3}+3 \times(2 p) \times \frac{1}{5}\left[2 p+\frac{1}{5}\right]$

Now, using $a^{3}+b^{3}+3 a b(a+b)=(a+b)^{3}$

$=\left(2 p+\frac{1}{5}\right)^{3}=\left(2 p+\frac{1}{5}\right)\left(2 p+\frac{1}{5}\right)\left(2 p+\frac{1}{5}\right)$