Suppose that I have known that $X$ and $Y$ have high dependency, $Y$ and $Z$ have high dependency, and $Z$ and $X$ also have high dependency through three different 2-copulas.
Suppose I fit one 3-copula, will my results be superior to the ones where I fitted three 2-copulas?
Having a 3-copula vs. three 2-copulas it is a bit like being able to view a 3D object in 3D from every angle vs. only being able to see three 2D shots of the object, each from a different angle (where a different dimension is completely hidden when seen from each of the three angles). The 2D shots tell you something but not everything about the 3D object.
From a 3-copula, you will be able to obtain the density $f_{X,Y,Z}(x,y,z)$ of any point $(x,y,z)$ and the probability $P(X\in R_X, Y\in R_Y, Z\in R_Z)$ of any region $(X\in R_X, Y\in R_Y, Z\in R_Z)$ for arbitrary regions $R_X$, $R_Y$ and $R_Z$. You will also be able to obtain conditional densities and probabilities such as $f_{X \mid Y,Z}(x \mid y,z)$ and $P(X\in R_X \mid Y\in R_Y, Z\in R_Z)$. You will not be able to obtain any of that from the three 2-copulas alone.
Whether you can call that superior or not depends on how you define superior. If you do not care about such densities and probabilities, then the 3-copula is probably not superior to the three 2-copulas. If you do, then it is.
