Question

Factorize:
1 + b³ + 8c³ − 6bc

Answer 1

$$\begin{gathered} 1+b^3+8c^3-6bc={\left(1\right)}^3+{\left(b\right)}^3+{\left(2c\right)}^3-3\times 1\times b\times 2c \\ =\left(1+b+2c\right)\left[1^2+b^2+{\left(2c\right)}^2-1\times b-b\times 2c-1\times 2c\right] \\ =\left(1+b+2c\right)\left(1+b^2+4c^2-b-2bc-2c\right) \end{gathered}$$