Consider the expression
![(x^3-1)/((x+1)^3+1)](https://img.qammunity.org/2022/formulas/mathematics/college/jwuwkckbqb7q2kpeku75nendclnjb89k8n.png)
Factorize the numerator and denominator a difference/sum of cubes:
![((x-1)(x^2+x+1))/(((x+1)+1)((x+1)^2-(x+1)+1))](https://img.qammunity.org/2022/formulas/mathematics/college/zn7snyfp8elaxqp8eqi5d8z1h13yqgjofl.png)
Expand the denominator:
![((x-1)(x^2+x+1))/((x+2)((x^2+2x+1)-(x+1)+1))=((x-1)(x^2+x+1))/((x+2)(x^2+x+1))=(x-1)/(x+2)](https://img.qammunity.org/2022/formulas/mathematics/college/jq4b8aaipto06uc8c3kfw8nb60ymkkrfy6.png)
since x = 2020, and clearly 2020² + 2020 + 1 ≠ 0, so we can cancel the factor of x ² + x + 1. This leaves us with
![(2020^3-1)/(2021^3+1) = (2020-1)/(2020+2) = (2019)/(2022)=(673)/(674)=\frac ab](https://img.qammunity.org/2022/formulas/mathematics/college/fjhn14014v67tq1w0jf0ipsf9471dcctq1.png)
so that a + b = 673 + 674 = 1347.