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The following is due to Lucas in 1876:

$$F_{n + 1}^3 + F_n^3 - F_{n - 1}^3 = F_{3n}$$

I am unable to locate an elementary proof of this identity, and am unable to reproduce it myself. Would anyone mind sharing a proof or a source?

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A nice strategy is to use the elementary identity $$ F_{n+m} = F_n F_{m-1} + F_{n+1}F_m. \tag1 $$ Use it to show that $\, a_n := F_{3n}\,$ satisfies the recursion $$ a_{n+2} = 4a_{n+1} + a_n. \tag2 $$ Now use it to express $\, F_{n-1} = F_{n+1} - F_n\,$ and hence express $$ b_n := F_{n + 1}^3 + F_n^3 - F_{n - 1}^3 \tag3 $$ in terms of $\, F_n\,$ and $\, F_{n+1}.\,$ Verify that $\,a_n\,$ and $\,b_n\,$ satisfy the same recursion and initial values. The idea is to express all of $\,b_n, b_{n+1}, b_{n+2}\,$ in terms of $\,F_n, F_{n+1}\,$ using simple algebra. $$ b_n = F_n(2F_n^2 -3F_nF_{n+1} +3F_{n+1}^2), \tag4 $$ $$ b_{n+1} = F_{n+1}(3F_n^2 +3F_nF_{n+1} +2F_{n+1}^2), \tag5 $$ $$ b_{n+2} = (F_n+F_{n+1})(2F_n^2 +7F_nF_{n+1} +8F_{n+1}^2). \tag6 $$

The OEIS sequence A014445 is defined as $a_n$ and its recursion as well as its equivalence to $b_n$ is given in the entry.

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  • $\begingroup$ Ok, the first relationship (2) can be shown with some playing around. Showing $b_n$ satisfies the same recursion seems to me to be roughly as difficult as the original problem, however. $\endgroup$ – Johnny Apple Nov 16 '18 at 4:25
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This is probably not the most elegant way, but you can use the explicit formula for the Fibonacci numbers. $$ F_n = \frac{1}{\sqrt{5}} \tau^n + \left(-\frac{1}{\sqrt{5}}\right)\left(-\frac{1}{\tau}\right)^n $$ Here $\tau = \frac{1}{2}+\frac{\sqrt{5}}{2}, -1/\tau = \frac{1}{2}-\frac{\sqrt{5}}{2}$. This formula for $F_n$ can now be plugged into the left-hand side of the desired identity. $$ \begin{align*} F_{n+1}^3 + F_n^3 - F_{n-1}^3 = & \left(\frac{1}{\sqrt{5}} \tau^{n+1} + \left(-\frac{1}{\sqrt{5}}\right)\left(-\frac{1}{\tau}\right)^{n+1}\right)^3\\ & + \left(\frac{1}{\sqrt{5}} \tau^n + \left(-\frac{1}{\sqrt{5}}\right)\left(-\frac{1}{\tau}\right)^n\right)^3\\ & - \left(\frac{1}{\sqrt{5}} \tau^{n-1} + \left(-\frac{1}{\sqrt{5}}\right)\left(-\frac{1}{\tau}\right)^{n-1}\right)^3 \end{align*}$$

Expanding out the terms takes a little work but will eventually lead to the desired expression $\frac{1}{\sqrt{5}} \tau^{3n} + \left(-\frac{1}{\sqrt{5}}\right)\left(-\frac{1}{\tau}\right)^{3n}$ for $F_{3n}$.

As an example, considering the leading terms of the three cubes yields $$ \frac{1}{5\sqrt{5}}\tau^{3n+3}+\frac{1}{5\sqrt{5}}\tau^{3n} - \frac{1}{5\sqrt{5}}\tau^{3n-3} = \frac{1}{5\sqrt{5}}\left(\tau^3 + 1 - \frac{1}{\tau^3}\right)\tau^{3n} $$ Since $\tau^3 = 2+ \sqrt{5}$ and $-1/\tau^3 = 2-\sqrt{5}$, we see that this simplifies to $(1/\sqrt{5})\tau^{3n}$, which is the first term of the expression for $F_{3n}$.

I leave the consideration of the other terms to the reader.

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Use Somos' first identity to expand $F_{2n+n}$ and then expand again to get a bunch of terms, each the product of three Fibonacci numbers.
Take out the common factor $F_n$, and find the factor $F_n$ in your left-hand side.
The rest should be fairly easy.

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