Factorise each of the following:
(i) 8a³ + b³ + 12ab + 6ab²
(ii) 8a³ - b³ - 12a²b + 6ab²
(iii) 27a³ + 225a²
(iv) 64a³ - 27b³ - 144a²b + 108ab²
(v) 27p³ - 1216p² + 14p

We know that
(a+b)³ = a³ + b³ + 3a²b + 3ab²
and (a-b)³ = a³ - b³ - 3a²b + 3ab²
(i) 8a³ + b³ + 12a²b + 6ab² = (2a)³ + (b)³ + 3(2a)²(b) + 3(2a)(b)² = (2a + b)³
(ii) 8a³ - b³ - 12a²b + 6ab² = (2a)³ - (b)³ - 3(2a)²(b) + 3(2a)(b)² = (2a - b)³
(iii) 27a³ + 225a² = This expression can't be factored using the formulas for the sum or difference of cubes. It can be factored by taking out a common factor:
27a³ + 225a² = 27a²(a + 25/3)
(iv) 64a³ - 27b³ - 144a²b + 108ab² = (4a)³ + (-3b)³ + 3(4a)²(-3b) + 3(4a)(-3b)² = (4a - 3b)³
(v) 27p³ - 1216p² + 14p = This expression cannot be easily factored using the sum or difference of cubes formulas. It requires different factoring techniques. Let's try factoring out a common factor of p:
27p³ - 1216p² + 14p = p(27p² - 1216p + 14)