Resolve `a^3-b^3+1+3a b` into factors.

Explanation

`a^3−b^3+1+3ab`
`=a^3+(−b)^3+1^3-3(axx(−b)xx1)`
`=(a+(−b)+1)(a^2+(−b)^2+1^2−a(−b)−(−b)(1)−1(a))`
`(x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2−xy−yz−zx))`
`=(a−b+1)(a^2+b^2+1+ab+b−a)`
Hence, `a^3−b^3+1+3ab=(a−b+1)(a^2+b^2+ab−a+b+1)`