Ttgjyuk

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?
$$F_nF_n-1F_n-2=fracF_n^3-F_n-1^3-F_n-2^33.$$

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

Remark. The motivation comes as follows. Define $F_n!=F_1cdots F_n$ and $F_0!=1$. Further, $binomnk_F:=fracF_n!F_k!cdot F_n-k!$. Then, I was studying these coefficients and was lead to
$$binomn3_F=fracF_n^3-F_n-1^3-F_n-2^33!.$$

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.

$F_n$ 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 $3F_n-1F_n-2^2$ 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
begineqnarray* F_n^3 & = & F_n-1^3 + F_n-2^3
+3(F_n-1^2F_n-2+F_n-1F_n-2^2)\ & = &
F_n-1^3 + F_n-2^3 +3F_n-1F_n-2(F_n-1+F_n-2)\ & = &
F_n-1^3 + F_n-2^3 + F_nF_n-1F_n-2. endeqnarray*