Why does Coulomb's Law hold for $F_{A(B,C)} = F_{AB} + F_{AC}$? Recently, my professor said that this fact was experimentally proven. I couldn't find it online, so hopefully, someone can either link it or give me a brief summary.
For three electric charges, $Q_A$, $Q_B$ and $Q_C$ that are positioned in space, it holds that the Coulomb Force between $Q_A$ and $Q_B$ is:
$$F_{AB} = k \frac{Q_A Q_B}{r_{AB}},$$
and similarly between $Q_A$ and $Q_C$:
$$F_{AC} = k \frac{Q_A Q_C}{r_{AC}}.$$
It then follows that the force $Q_B$ and $Q_C$ exert together on $Q_A$ is given as:
$$F_{A(B,C)} = F_{AB} + F_{AC}.$$
What confuses me about this is that although it seems logical, it's still something that could and doesn't have to hold for different cases.
 A: This is just called the principle of superposition. Electric fields and electric forces have been shown to add linearly, and so the net force experienced by charge A is equal to the sum of the forces individually acting on A.
Note that not all electric properties follow this. For example, the energy stored in an electric field involves the integral of the square of the field over all space: $U\propto\iiint E^2\,\text dV$, and so the energy stored in a linear combination of fields is not the sum of the individual energies of each field.
A: It is simply an empirical fact about classical electrostatic (or gravitational) forces that the there are only two-body forces.  That is, the force on a particle $m_{1}$, due to other particles $m_{2},m_{3},m_{4},\ldots$ is just a superposition
$$\vec{F}_{1}=\vec{F}_{12}+\vec{F}_{13}+\vec{F}_{14}+\cdots,$$
where each $\vec{F}_{1j}$ is the force that would exist between $m_{1}$ and $m_{j}$ if there were no other particles present. (Note that these are vector equations; the forces add as three-dimensional vectors, not just as magnitudes.)
This is, obviously, the simplest form that a force law can take when there are more than two interacting bodies, and there are reasons why this kind of simple force law emerges as the classical, nonrelativistic limit of a more fundamental theory like quantum electrodynamics.  However, when Coulomb's Law describing the forces between stationary charges was being measured and verified, that fact that there were no additional three-body and higher forces was simply something that was observed to be true experimentally.  There is, classically, no way to derive this fact; it is just an experimental observation.
The other important thing about the superposition law $\vec{F}_{1}=\vec{F}_{12}+\vec{F}_{13}+\vec{F}_{14}+\cdots,$ is that, in spite of its simplicity, it did not need to be true.  In more complicated systems, there actually are irreducible three-body forces—meaning that the force on a body due to two or more others cannot be written as this kind of simple sum of two-body forces.  An extremely important example of this comes from nuclear physics:  the forces between multiple helium nuclei (or $\alpha$-particles) $^{4}$He$^{2+}$.  The charges on the nuclei create a repulsive Coulomb force that dominates the force between two helium nuclei at large distances.  However, at very short distances (on the scale of $\sim10^{-15}$ m) there is also a short-range attractive force.  This attractive nuclear force is responsible for the fact that in a collision of two $^{4}$He$^{2+}$ nuclei, they may temporarily be pulled together to form a $^{8}$Be$^{4+}$.  However, ultimately, the attractive nuclear force between the two $^{4}$He$^{2+}$ is not strong enough to win out over the repulsive electrostatic force; this makes $^{8}$Be$^{4+}$ unstable, and it quickly falls apart into two $\alpha$-particles again.  However, if you add a third $^{4}$He$^{2+}$ nucleus to the system, there are a intrinsically three-body forces that, at short ranges, are attractive enough to hold together three $\alpha$-particles, fusing making a stable $^{12}$C$^{6+}$ nucleus.  Since carbon is one of the key building blocks of life, we would not be here at all if it were not for these three-body forces.
A: The force exerted on an object is a superposition of forces from another objects. It is true for different nature of forces. By the way in your case you should write as following:
\begin{equation}
\vec{F}_{AB} = k \frac{Q_A Q_B}{(r_{AB})^3}\cdot\vec{r}_{AB},
\end{equation}
\begin{equation}
\vec{F}_{AC} = k \frac{Q_A Q_B}{(r_{AC})^3}\cdot\vec{r}_{AC},
\end{equation}
Attention, the force is to the power -2 and not -1 as you wrote
And the superpostion
\begin{equation}
\vec{F}_{A(B,C)} = \vec{F}_{AB} + \vec{F}_{AC}
\end{equation}
