[tex]\cosh ax+\sinh ax=e^{ax}=(e^x)^a=(\cosh x+\sinh x)^a[/tex]
[tex]\cosh x=\cos ix[/tex]
[tex]\sinh x=-i\sin ix[/tex]
By DeMoivre's theorem,
[tex]\cos3ix-\sin3ix=(\cos ix-i\sin ix)^3=\cos^3ix-3i\cos^2ix\sin ix-3\cos ix\sin^2ix+i\sin^3ix[/tex]
You then have
[tex]\cos3ix=\mathrm{Re}\left[(\cos ix-i\sin ix)^3\right][/tex]
[tex]\cos3ix=\cos^3ix-3\cos ix\sin^2ix[/tex]
[tex]\cos3ix=\cos^3ix+3\cos ix(-i\sin ix)^2[/tex]
Converting back to hyperbolic functions, you get
[tex]\cosh3x=\cosh^3x+3\cosh x\sinh^2x[/tex]
