Factorise

$1+\frac{27}{125} a^{3}+\frac{9 a}{5}+\frac{27 a^{2}}{25}=(1)^{3}+\left(\frac{3}{5} a\right)^{3}+3(1)^{2}\left(\frac{3}{5} a\right)+3(1)\left(\frac{3}{5} a\right)^{2}$

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

Hence, factorisation of $1+\frac{27}{125} a^{3}+\frac{9 a}{5}+\frac{27 a^{2}}{25}$ is $\left(1+\frac{3}{5} a\right)^{3}$.