Factorize:

1 + b³ + 8c³ – 6bc

We know that,

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)]

Using this formula, we get

= (1)³ + (b)³ + (2c)³ – 3 (1) (b) (2c)

= (1 + b + 2c) [(1)² + (b)² + (4c)² – (1) (b) – (2b) (c) – (2c) (1)]

= (1 + b + 2c) (1 + b² + 4c² – b – 2bc – 2c)