Everywhere I've looked so far (such as NIST) the Fermi coupling constant $G_F$ is always expressed as
$$\frac{G_F}{(\hbar c)^3} = 1.166 364(5) \times 10^{-5} \textrm{ GeV}^{-2}$$
never as just plain old $G_F$. I am wondering why that is.
Answer
This is mostly to make an explicit connection with natural units - the unit system in which $\hbar$ and $c$ are both set to 1, which is the natural set of units for relativistic quantum theory. Because you've adimensionalized two units and you had three physical dimensions to start with (mass, length and time), natural units retain a single dimensional parameter, which is usually taken to be mass and, because this is usually particle physics we're talking about, measured in $\mathrm{eV}/c^2$, or just $\mathrm{eV}$ with the factor of $c=1$ understood.
Physical quantities in natural units therefore always carry a single physical dimension, which can always be expressed in terms of a power of mass, and this power is known as the mass dimension of the quantity. Time, for example, has dimensions of $M^{-1}$, as does length. The Fermi constant has mass dimension of -2, so in natural units it has units of $\mathrm{eV}^{-2}$.
The expression you give has the correct powers of $\hbar$ and $c$ such that $G_F$ will have the correct dimensionality in standard systems of units, but it keeps these factors explicitly so that the numerical value will be conserved if one goes into natural units. This is exactly analogous to reporting a mass in $\mathrm{eV}/c^2$: formally correct in SI units, gives directly the value in natural units, and lets one focus on the scales that one wants to focus on without any unit-conversion hassle.
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