electrostatics - Why do we say that in Coulomb's law the force is proportional to $frac{1}{r^{2}}$ and not $frac{1}{r^{3}}$?

I am going over Coulomb's law and there is something that is a bit confusing for me:

According to Coulomb's law, if I have a charge $q_{1}$ at a point $\vec{r_{1}}$ and a charge $q_{2}$ at a point $\vec{r_{2}}$ then the force that the first charge applies to the second charge is given by $$ F_{1,2}=\frac{Kq_{1}q_{2}}{|\vec{r_{2}}-\vec{r_{1}}|^{2}}(\vec{r_{2}}-\vec{r_{1}})=\frac{Kq_{1}q_{2}}{|\vec{r_{2}}-\vec{r_{1}}|^{3}}\hat{(\vec{r_{2}}-\vec{r_{1}})} $$

I see that in the first expression is does look like the force is proportional to $\frac{1}{r^{2}}$, but I don't understand why it isn't the second expression that matters, giving us that the force is proportional to $\frac{1}{r^{3}}$:

Say both charges are on the $x-$axis, then $\hat{(\vec{r_{2}}-\vec{r_{1}})}=\hat{x}$. Increasing the distance between the charges to two times what is by moving the second charge on the $x$- axis was, would decrease the force $2^{3}=8$ times, since $\hat{(\vec{r_{2}}-\vec{r_{1}})}$ is still $\hat{x}$.

What is the mistake in my reasoning ?

Answer

The mistake you made is in the way you stated Coloumb's law.

It's either

$$ \vec{F} = K \frac{q_1 q_2}{r^\color{red}3} \color{red}{\vec{r}} $$ OR $$ \vec{F} = K \frac{q_1 q_2}{r^\color{red}2} \color{red}{\hat{r}} $$

but definitely NOT

$$ \vec{F} = K \frac{q_1 q_2}{r^\color{red}3} \color{red}{\hat{r}} $$