Products and sum of cubes in Fibonacci

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 $3F_{n-1}F_{n-2}^2$ 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*}