Factorize:

$1+b^{3}+8 c^{3}-6 b c=(1)^{3}+(b)^{3}+(2 c)^{3}-3 \times 1 \times b \times 2 c$

$=(1+b+2 c)\left[1^{2}+b^{2}+(2 c)^{2}-1 \times b-b \times 2 c-1 \times 2 c\right]$

$=(1+b+2 c)\left(1+b^{2}+4 c^{2}-b-2 b c-2 c\right)$