Magnitude of fourier transform of AM signal
Your computation of the cross-term is wrong (at least it was, before you edited). The squared magnitude of the amplitude modulated spectrum is

$$|X_{AM}(f)|^2=|X(f-f_c)|^2+|X(f+f_c)|^2+2|X(f-f_c)||X(f+f_c|\cos(\Delta\phi)\tag{1}$$

where \$\Delta\phi\$ is the phase difference between \$X(f-f_c)\$ and \$X(f+f_c)\$. If \$f_c&gt;f_m\$ (which is usually the case in practice), the two shifted spectra do not overlap and, consequently, the cross-term is zero. For taking the square root a similar reasoning applies: since \$X(f-f_c)\$ and \$X(f+f_c)\$ do not overlap, the square root of the sum is simply the sum of the square roots. Consequently, you get

$$|X_{AM}(f)|=|X(f-f_c)|+|X(f+f_c)|\tag{2}$$