Find
$$\sum _ { r = 1 } ^ { n } \left( r ^ { 3 } - 6 r \right)$$
expressing your answer in the form
$$k n ( n + 1 ) ( n + p ) ( n + q )$$
where \(k\) is a fraction and \(p\) and \(q\) are integers.
It is given that
$$S = \sum _ { r = 1 } ^ { 1000 } \left( r ^ { 3 } - 6 r \right)$$
Without calculating the value of \(S\), show that \(S\) is a multiple of 2008 .