We know that, a 3 + b 3 + c 3 – 3abc = (a + b + c) (a 2 + b 2 + c 2 – ab – bc – ca)] Using this formula, we get = ( a ) 3 + (2 b) 3 + (c) 3 – 3 ( a ) (2 b) (c) = ( + 2 b + c) [( ) 2 + (2 b) 2 + (c) 2 – ( ) (2 b) – (2 b) (c) – (c) ( a )] = ( + 2 b + c) (2a 2 + 8b 2 + c 2 – 4ab – 2 bc – ac)