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