Consider the familiar sequence of [Fibonacci numbers](https://en.wikipedia.org/wiki/Fibonacci_number): $F\_0=0, F\_1=1, F\_n=F\_{n-1}+F\_{n-2}$.


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
$$\binom{n}3\_F=\frac{F\_n^3-F\_{n-1}^3-F\_{n-2}^3}{3!}.$$



Products and sum of cubes in Fibonacci
$F\_n$ is the number of compositions (ordered partitions) of $n-1$ into
parts equal to 1 or 2. The number of triples $(a,b,c)$ of such
compositions is $F\_n^3$. The number such that $a,b,c$ all begin with 1
is $F\_{n-1}^3$. The number such that $a,b,c$ all begin with 2 is
$F\_{n-2}^3$. Otherwise either one of $a,b,c$ begins with 1 and the
others begin with 2, or vice versa. There are $3F\_{n-1}F\_{n-2}^2$ such
triples of the first type. Similarly there are $3F\_{n-1}^2F\_{n-2}$
of the second type, i.e., one of
$a,b,c$ begins with 2 and the others begin with 1. Hence
 \begin{eqnarray\*} F\_n^3 & = & F\_{n-1}^3 + F\_{n-2}^3
 +3(F\_{n-1}^2F\_{n-2}+F\_{n-1}F\_{n-2}^2)\\ & = &
 F\_{n-1}^3 + F\_{n-2}^3 +3F\_{n-1}F\_{n-2}(F\_{n-1}+F\_{n-2})\\ & = &
 F\_{n-1}^3 + F\_{n-2}^3 + 3F\_nF\_{n-1}F\_{n-2}. \end{eqnarray\*}


