Factorise the following

(i) $1-64 a^{3}-12 a+48 a^{2}=(1)^{3}-(4 a)^{3}-3 \times 1^{2} \times 4 a+3 \times 1 \times(4 a)^{2}$

[using identity, $(a-b)^{3}=a^{3}-b^{3}-3 a^{2} b+3 a b^{2}$ ]

$=(1-4 a)^{3}=(1-4 a)(1-4 a)(1-4 a)$

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

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

[using identity, $(a+b)^{3}=a^{3}+b^{3}+3 a^{2} b+3 a b^{2}$ ]

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