Question
Factorise: $216 + 27b^3 + 8c^3 - 108bc$
✓
Answer
$216 + 27b^3 + 8c^3 - 108bc$
$= (6)^3 + (3b)^3 + (2c)^3 - 3 \times 6 \times 3b \times 2c$
$= (6 + 3b + 2c)[6^2 + (3b)^2 + (2c)^2 - 6 \times 3b - 3b \times 2c - 2c \times 6]$
$= (6 + 3b + 2c)(36 + 9b^2 + 4c^2 - 18b - 6bc - 12c)$
$= (6)^3 + (3b)^3 + (2c)^3 - 3 \times 6 \times 3b \times 2c$
$= (6 + 3b + 2c)[6^2 + (3b)^2 + (2c)^2 - 6 \times 3b - 3b \times 2c - 2c \times 6]$
$= (6 + 3b + 2c)(36 + 9b^2 + 4c^2 - 18b - 6bc - 12c)$
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