An algebra problem by Sakanksha Deo

In general, the sum of the cubes of the first $n$ odd positive integers is

$\displaystyle\sum_{k=1}^{n} (2k - 1)^{3} = \sum_{k=1}^{n} (8k^{3} - 12k^{2} + 6k - 1) =$

$\displaystyle 8\sum_{k=1}^{n} k^{3} - 12\sum_{k=1}^{n} k^{2} + 6\sum_{k=1}^{n} k - \sum_{k=1}^{n} 1 =$

$8*\left(\dfrac{n(n + 1)}{2}\right)^{2} - 12*\dfrac{n(n + 1)(2n + 1)}{6} + 6*\dfrac{n(n + 1)}{2} - n =$

$2n^{2}(n + 1)^{2} - 2n(n + 1)(2n + 1) + 3n(n + 1) - n =$

$n(n + 1)[2n(n + 1) - 2(2n + 1) + 3] - n =$

$n(n + 1)(2n^{2} - 2n + 1) - n = n(2n^{3} - n + 1) - n = n^{2}(2n^{2} - 1).$

Plugging in $n = 500$ gives us a solution of

$500^{2}(2*500^{2} - 1) = \boxed{124999750000}.$