5 Use the standard results for \(\sum _ { r = 1 } ^ { n } r , \sum _ { r = 1 } ^ { n } r ^ { 2 }\) and \(\sum _ { r = 1 } ^ { n } r ^ { 3 }\) to show that, for all positive integers \(n\),
$$\sum _ { r = 1 } ^ { n } \left( 8 r ^ { 3 } - 6 r ^ { 2 } + 2 r \right) = 2 n ^ { 3 } ( n + 1 )$$