Products and sum of cubes in Fibonacci

Question

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 $$\binom{n}3_F=\frac{F_n^3-F_{n-1}^3-F_{n-2}^3}{3!}.$$

Answer 1

$F_n$ 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*}

Answer 2

This is just the following identity: $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca).$$ Since $$F_n+(-F_{n-1})+(-F_{n-2})=0,$$ your formula follows.