Show that
$$\cosh ^ { 3 } x + \sinh ^ { 3 } x = \frac { 1 } { 4 } \mathrm { e } ^ { m x } + \frac { 3 } { 4 } \mathrm { e } ^ { n x }$$
where \(m\) and \(n\) are integers.
6
Hence find \(\cosh ^ { 6 } x - \sinh ^ { 6 } x\) in the form
$$\frac { a \cosh ( k x ) + b } { 8 }$$
where \(a , b\) and \(k\) are integers.