Consider the expression
[tex]\dfrac{x^3-1}{(x+1)^3+1}[/tex]
Factorize the numerator and denominator a difference/sum of cubes:
[tex]\dfrac{(x-1)(x^2+x+1)}{((x+1)+1)((x+1)^2-(x+1)+1)}[/tex]
Expand the denominator:
[tex]\dfrac{(x-1)(x^2+x+1)}{(x+2)((x^2+2x+1)-(x+1)+1)}=\dfrac{(x-1)(x^2+x+1)}{(x+2)(x^2+x+1)}=\dfrac{x-1}{x+2}[/tex]
since x = 2020, and clearly 2020² + 2020 + 1 ≠ 0, so we can cancel the factor of x ² + x + 1. This leaves us with
[tex]\dfrac{2020^3-1}{2021^3+1} = \dfrac{2020-1}{2020+2} = \dfrac{2019}{2022}=\dfrac{673}{674}=\dfrac ab[/tex]
so that a + b = 673 + 674 = 1347.
