<s>[INST] Fibonacci Cubes: $F_n^3 + F_{n+1}^3 - F_{n-1}^3 =F_{3n}$ 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. 
 [/INST] Here you'll sort of see my thought process; hopefully it helps.
First you need to define the Fibonacci sequence as $F_1 = 1, F_2 = 1, F_{n+1} = F_{n} + F_{n-1}$ for this to hold. If you somehow want to use induction, you need to write $F_{n+1}$ with the recurrence relation, whence $S = F_{n}^3 - F_{n-1}^3 + (F_{n}+F_{n-1})^3 = F_{n}^3 - F_{n-1}^3 + 3F_n^2 F_{n-1} + 3F_n F_{n-1}^2 + F_{n}^3 + F_{n-1}^3.$ 
Writing $-F_{n-2} = F_{n-1} - F_{n}$ gives $-F_{n-2}^3 = F_{n-1}^3 - F_{n}^3 + 3F_{n}^2F_{n-1} - 3F_n F_{n-1}^2$ and thus, $S = F_n^3 + F_{n-1}^3 - F_{n-2}^3 + 2(F_n^3 - F_{n-1}^3 + 3F_n F_{n-1}^2) = F_{3n-3}+2(F_n^3 - F_{n-1}^3 + 3F_n F_{n-1}^2).$ 
To make it clear what exactly we need, if $S = F_{3n},$ then $S = F_{3n-1} + F_{3n-2} = 2F_{3n-2} + F_{3n-3},$ so in fact we want to prove that $F_{n}^3 + 3F_{n}F_{n-1}^2 - F_{n-1}^3 = F_{3n-2}.$ Looking back at our first equation for $S,$ this means we also need $F_{n}^3 + 3F_{n}^2F_{n-1} + F_{n-1}^3 = F_{3n-1}.$ 
Claim: $F_{n}^3 + 3F_{n}F_{n-1}^2 - F_{n-1}^3 = F_{3n-2}$ and $F_{n}^3 + 3F_{n}^2F_{n-1} + F_{n-1}^3 = F_{3n-1}.$
Proof: The base cases are easy to check. We proceed by double induction. That is, assume the claim holds for some $n.$ Then for $n+1,$ we have 
$\begin{align*}
F_{n+1}^3 + 3F_{n+1}F_{n}^2 - F_{n}^3& = (F_n + F_{n-1})^3 + 3F_n^3 + 3F_n^2 F_{n-1} - F_n^3
\\&= \underbrace{F_{n}^3 + 3F_{n}^2 F_{n-1}} + 3F_{n}F_{n-1}^2 + \underbrace{F_{n-1}^3} + 2F_{n}^3 + 3F_{n}^2F_{n-1} \\&= F_{3n-1} + 2F_{n}^3 + 3F_{n}^2F_{n-1} + 3F_{n}F^2_{n-1}\\
&= F_{3n-1} + (F_n^3 + 3F_{n}^2F_{n-1} + F_{n-1}^3) + (F_{n}^3+3F_{n}F_{n-1}^2 - F_{n-1}^3)\\
&=F_{3n-1} + F_{3n-1} + F_{3n-2} = F_{3n-1} + F_{3n} = F_{3n+1} = F_{3(n+1) - 2}.
\end{align*}$
Similarly, one shows that $F_{n+1}^3 + 3F_{n+1}^2 F_n + F_{n}^3 = F_{3(n+1) - 1},$ whence we are done.
tl;dr: Abuse the Fibonacci recurrence relation and the inductive hypotheses.
</s>