(i) (a) 3 + (−2b) 3 + (−4c) 3 − 3(a)(−2b)(−4c) =[a + (−2b + ( – 4c)][(a) 2 + (−2b) 2 + (−4c) 2 −(a)(−2b)−(−2b)(−4c)− (−4c)(a)] [ ∵ a 3 + b 3 + c 3 – 3abc = (a + b + c)(a 2 + b 2 + c 2 – ab – bc – ca) ] = (a– 2b – 4c)(a 2 + 4b 2 + 16c 2 + 2ab – 8bc + 4ca) (ii). Given that: 2√2a 3 + 8b 3 – 27c 3 + 18√2abc = (√2a) 3 + (2b) 3 + (−3c) 3 − 3(√2a)(2b)(3c) =[√2a + (2b) + (−3c)][(√2a) 2 + (2b) 2 + (−3c) 2 −(√2a)(2b)−(2b)(−3c)− (−3c)(√2a)] [ ∵ a 3 + b 3 + c 3 – 3abc = (a + b + c)(a 2 + b 2 + c 2 – ab – bc – ca) ] = (√2a + 2b – 3c)(2a 2 + 4b 2 + 9c 2 − 2√2ab + 6bc + 3√2ca)

Factorise: (i).

Factorise:
(i).

(i) (a)³ + (−2b)³ + (−4c)³ − 3(a)(−2b)(−4c)

=[a + (−2b + ( – 4c)][(a)² + (−2b)² + (−4c)²−(a)(−2b)−(−2b)(−4c)− (−4c)(a)]

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

= (a– 2b – 4c)(a² + 4b² + 16c² + 2ab – 8bc + 4ca)

(ii). Given that: 2√2a³ + 8b³ – 27c³ + 18√2abc

= (√2a)³ + (2b)³ + (−3c)³ − 3(√2a)(2b)(3c)

=[√2a + (2b) + (−3c)][(√2a)² + (2b)² + (−3c)²−(√2a)(2b)−(2b)(−3c)− (−3c)(√2a)]

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

= (√2a + 2b – 3c)(2a² + 4b² + 9c² − 2√2ab + 6bc + 3√2ca)

Factorise:(i).

Factorise:
(i).