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

Explanation

We factorize the given expression` 'a³ - b³ +1+3ab' `as shown below:
`a³-b³+1+3ab`
=`a³+ (-b)³+1³+ 3(a × b × 1)`
= `(a + (-b) + 1)(a² + (-b)² + 1² - a(-b) - (-b)(1) -1(a))`
`[(x³+y³+z³+3abc = (x+y+z)(x² + y² + z² - xy −yz−zx)]`
= `(a - b + 1)(a² + b² + 1 + ab + b - a)`
Hence, `a³-b³+1+3ab = (a - b + 1)(a + b² + ab - a +b+1)`