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

• I get a different left hand side. Rewrite the n+2 term as the sum of n+1 and n terms, and then compute the difference of cubes and divide by three. Algebraically you get the product of the n term and the n+1 term and (the sum of n+1 and n terms). This seems to have more to do with (a+b)^n - a^n - b^n than with Fibonacci. Gerhard "Unsure Of Any Combinatorial Interpretation" Paseman, 2019.03.26. – Gerhard Paseman Mar 26 at 18:33
• Thanks, edited accordingly. – T. Amdeberhan Mar 26 at 18:40
• $(a+b)^3 - a^3 - b^3 = 3ab(a+b)$. If $a,b$ are consecutive Fibonacci numbers then $a+b$ is the next. – Noam D. Elkies Mar 26 at 19:58

$$F_n$$ 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 $$\begin{eqnarray*} F_n^3 & = & F_{n-1}^3 + F_{n-2}^3 +3(F_{n-1}^2F_{n-2}+F_{n-1}F_{n-2}^2)\\ & = & F_{n-1}^3 + F_{n-2}^3 +3F_{n-1}F_{n-2}(F_{n-1}+F_{n-2})\\ & = & F_{n-1}^3 + F_{n-2}^3 + 3F_nF_{n-1}F_{n-2}. \end{eqnarray*}$$
• The OP specified a desire for "combinatorial" or "conceptual" explanations. But the distinction between combinatorics and algebra is blurry here. You have to choose one ball from each of three urns, each containing $a$ amaranth and $b$ blue balls. How many choices don't have all three balls the same color? On the one hand, $(a+b)^3-a^3-b^3$. On the other, choose any cyclic permutation of (amaranth, blue, either) to get $3ab(a+b)$. – Noam D. Elkies Mar 27 at 0:14
• This generalizes to give $F_{n-1} F_{n-2} = (F_n^2 - F_{n-1}^2 - F_{n-2}^2)/2$ (by considering pairs of compositions) but one doesn't get anything similarly nice for fourth powers as far as I can tell. – Michael Lugo Mar 27 at 14:22
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.
• Simpler yet: since $F_n = F_{n-1} + F_{n+2}$ it's enough to use the two-variable identity $(a+b)^3 - a^3 - b^3 = 3ab(a+b)$ which is a quick consequence of the binomial expansion of $(a+b)^3$. – Noam D. Elkies Mar 26 at 20:00