Question
Factorise:
27a3 - b3 + 8c3 - 18abc
27a3 - b3 + 8c3 - 18abc
✓
Answer
27a3 - b3 + 8c3 - 18abc
= (3a)3 + (-b)3 + (2c)3 - 3 × (3a) × (-b) × (2c)
= [3a + (-b) + 2c][(3a)2 + (-b)2 + (2c)2 - 3a(-b)2c - 3a × 2c]
= (3a - b + 2c)(9a2 + b2 + 4c2 + 3ab + 2bc - 6ac)
= (3a)3 + (-b)3 + (2c)3 - 3 × (3a) × (-b) × (2c)
= [3a + (-b) + 2c][(3a)2 + (-b)2 + (2c)2 - 3a(-b)2c - 3a × 2c]
= (3a - b + 2c)(9a2 + b2 + 4c2 + 3ab + 2bc - 6ac)
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