Products and sum of cubes in Fibonacci

Consider the familiar sequence of Fibonacci numbers: $ F_0=0, F_1=1, F_n=F_{n-1}+F_{n-2}$ .

Although it is rather easy to furnish an algebraic verification of the below identity, I wish to see a different approach. Hence,

QUESTION. Is there a combinatorial or more conceptual reason for this “pretty” identity? $ $ F_nF_{n-1}F_{n-2}=\frac{F_n^3-F_{n-1}^3-F_{n-2}^3}3.$ $

Caveat. I’m open to as many alternative replies, of course.

Remark. The motivation comes as follows. Define $ F_n!=F_1\cdots F_n$ and $ F_0!=1$ . Further, $ \binom{n}k_F:=\frac{F_n!}{F_k!\cdot F_{n-k}!}$ . Then, I was studying these coefficients and was lead to $ $ \binom{n}3_F=\frac{F_n^3-F_{n-1}^3-F_{n-2}^3}{3!}.$ $