Factorize:

216 + 27b3 + 8c3 – 108bc


We know that,


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


Using this formula, we get


= (6)3 + (3b)3 + (2c)3 – 3 (6) (3b) (2c)


= (6 + 3b + 2c) [(6)2 + (3b)2 + (2c)2 – (6) (3b) – (3b) (2c) – (2c) (6)]


= (6 + 3b + 2c) (36 + 9b2 + 4c2 – 18ab – 6bc – 12ac)


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