Question
Factorise:
$1+b^3+8 c^3-6 b c$

Answer

$1+b^3+8 c^3-6 b c$
$=(1)^3+(b)^3+(2 c)^3-3 \times 1 \times b \times 2 c$
$=(1+b+2 c)\left[1^2+b^2+(2 c)^2-1 \times b-b \times 2 c-1 \times 2 c\right]$
$=(1+b+2 c)\left(1^2+b^2+4 c^2-b-2 b c-2 c\right)$

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