Factorise `a^3-b^3+1+3ab`.

Explanation

We have
`a^3-b^3+1+3ab`
`=a^3+(-b)^3+(1)^3-3xx a xx (-b)xx 1`
`=x^3+y^3+z^3-3xyz, " where" a =x, (-b) =y and 1=z`
`=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)`
`=(a-b+1)(a^2+b^2+1+ab+b-a)`
`=(a-b+1)(a^2+b^2+ab-a+b1)`.