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)