Factorise:

$9 a^{2}+6 a+1-36 b^{2}$

$=\left[(3 a)^{2}+2 \times 3 a \times 1+1^{2}\right]-(6 b)^{2}$

$=(3 a+1)^{2}-(6 b)^{2} \quad\left[a^{2}+2 a b+b^{2}=(a+b)^{2}\right]$

$=(3 a+1-6 b)(3 a+1+6 b) \quad\left[a^{2}-b^{2}=(a-b)(a+b)\right]$

$=(3 a-6 b+1)(3 a+6 b+1)$