`sum_(n = 1)^20 (n^3 - n)` Find the sum using formulas for the sums of powers of integers.

You need to evaluate the given sum using formulas for the sums of powers of integers, such that:

`sigma_(n=1)^20 (n^3 - n) = sigma_(n=1)^20 n^3 - sigma_(n=1)^20 n`

`sigma_(n=1)^20 n^3 = 1^3 + 2^3 + .... + 20^3`

You need to remember that `sigma_(n=1)^n k^3 = ((n(n+1))/2)^2`

Replacing 20 for n yields:

`sigma_(n=1)^20 n^3 = ((20(20+1))/2)^2`

`sigma_(n=1)^20 n^3 = 100*21^2`

`sigma_(n=1)^20 n = 1 + 2 + ... + 20`

You need to remember that `sigma_(n=1)^n k = ((n(n+1))/2)`

Replacing 20 for n yields:

`sigma_(n=1)^20 n = (20(20+1))/2`

`sigma_(n=1)^20 n = 10*21`

Evaluating the sum, yields:

`sigma_(n=1)^20 (n^3 - n) = 100*21^2 - 10*21`

Factoring out `10*21` yields:

`sigma_(n=1)^20 (n^3 - n) = 10*21*(10*21 - 1)`

`sigma_(n=1)^20 (n^3 - n) = 210*209`

`sigma_(n=1)^20 (n^3 - n) = 43890`

Hence, evaluating the given sum using formulas for the sums of powers of integers, yields `sigma_(n=1)^20 (n^3 - n) = 210*209 =  43890.`