Verify that:

27a³ + b³ + c³ – 9abc = (3a + b + c)[3a² + b² + c² – 3ab – bc – 3ac]

Taking L.H.S

27a³ + b³ + c³ – 9abc = (3a)³ + b³ + c³ – 3(3a)(b)(c)

Using identity:

x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z^{2 –} xy – yz – zx)

⇒ 27a³ + b³ + c³ – 9abc = (3a + b + c)((3a)² + b² + c² – 3ab – bc – c(3a))

= (3a + b + c)[3a² + b² + c² – 3ab – bc – 3ac]

= R.H.S