Question
Factorise:
1 + b3 + 8c3 - 6bc
1 + b3 + 8c3 - 6bc
✓
Answer
1 + b3 + 8c3 - 6bc
= (1)3 + (b)3 + (2c)3 - 3 × 1 × b × 2c
= (1 + b + 2c)[12 + b2 + (2c)2 - 1 × b - b × 2c - 1 × 2c]
= (1 + b + 2c)(12 + b2 + 4c2 - b - 2bc - 2c)
= (1)3 + (b)3 + (2c)3 - 3 × 1 × b × 2c
= (1 + b + 2c)[12 + b2 + (2c)2 - 1 × b - b × 2c - 1 × 2c]
= (1 + b + 2c)(12 + b2 + 4c2 - b - 2bc - 2c)
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