TITLE: Fibonacci Cubes: $F_n^3 + F_{n+1}^3 - F_{n-1}^3 =F_{3n}$
QUESTION [11 upvotes]: Prove 
$$F_n^3 + F_{n+1}^3 - F_{n-1}^3 =F_{3n}$$
I've tried induction, either it's just very long or a neat trick is required in the inductive step but for some odd reason it's not working out. Ideally I would like any suggestions for the inductive proof.

REPLY [3 votes]: Maybe you could start by showing that
$$F_{a+b}=F_aF_{b+1}+F_{a-1}F_b. \tag{1}$$
IIRC this is sometimes called convolution identity.
One possible derivation of this identity is given on Wikipedia.
(I'll add also a link to the recent revision of the Wikipedia article - just in the case it changes in the future.)
See also this post: Showing that an equation holds true with a Fibonacci sequence: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$
Very similar identity is shown in this post: Proof of identity $F_m F_n + F_{m−1} F_{n−1} = F_{m+n−1}$ for Fibonacci numbers
Using (1) for $a=b=n$ you get
$$F_{2n}=F_n(F_{n+1}+F_{n-1})$$
Using (1) for  $a=n+1$ and $b=n$ you get
$$F_{2n+1}=F_{n+1}^2+F_n^2.$$
Now if we use (1) for $a=n$ and $b=2n$, we obtain
\begin{align}
F_{3n}&=F_nF_{2n+1}+F_{n-1}F_{2n}=\\
&=F_n(F_{n+1}^2+F_n^2)+F_{n-1}F_n(F_{n+1}+F_{n-1})=\\
&=F_n^3+F_nF_{n+1}^2+F_{n-1}(F_{n+1}-F_{n-1})(F_{n+1}+F_{n-1})=\\
&=F_n^3+F_nF_{n+1}^2+F_{n-1}(F_{n+1}^2-F_{n-1}^2)=\\
&=F_n^3+F_nF_{n+1}^2+F_{n-1}F_{n+1}^2-F_{n-1}^3=\\
&=F_n^3+(F_n+F_{n-1})F_{n+1}^2-F_{n-1}^3=\\
&=F_n^3+F_{n+1}^3-F_{n-1}^3
\end{align}