Question

Factorise: a³ − b³ + 1 + 3ab.

Answer 1

$$\begin{gathered} a^3-b^3+1+3ab \\ =a^3+{\left(-b\right)}^3+1^3-3a\left(-b\right)1 \\ =\left[a+\left(-b\right)+1\right]\left[a^2+{\left(-b\right)}^2+1^2-a\left(-b\right)-\left(-b\right)1-1a\right] \\ =\left(a-b+1\right)\left(a^2+b^2+1+ab+b-a\right) \end{gathered}$$