Show that \(( 1 + \mathrm { i } ) ^ { 3 } = 2 \mathrm { i } - 2\).
The cubic equation
$$z ^ { 3 } - ( 5 + \mathrm { i } ) z ^ { 2 } + ( 9 + 4 \mathrm { i } ) z + k ( 1 + \mathrm { i } ) = 0$$
where \(k\) is a real constant, has roots \(\alpha , \beta\) and \(\gamma\).
It is given that \(\alpha = 1 + \mathrm { i }\).