<s>[INST] Show that $S_L= L_n^2L_{n+1} -L_1L_2L_3-L_2L_3L_4-\cdots L_{n-2}L_{n-1}L_n$, $n\ge3$ $2,1,3,4,7,11,...$ for $n=0,1,2,3,4,...$; It is the Lucas numbers
Let the sum of the cube of Lucas series be $S_L$
$n\ge3$
$S_L=2^3+1^3+3^3+4^3+\cdots+L_n^3$,
show that it has a form of $S_L=L_n^2L_{n+1}-L_1L_2L_3-L_2L_3L_4-\cdots-L_{n-2}L_{n-1}L_n$
I try: I can't think  of any simple identities to use.
This is the only one might have some sort of link to it,
$L_{n+1}^3+L_{n+2}^3={L_{n+3}\over 2}(L_n^2+L_{n+1}^2+L_{n+2}^2)$
Any further hints?
 [/INST] In the given form, we can easily prove it via induction. Verify the case $n = 3$ by hand, and for $n \geqslant 4$, use the recurrence to get
\begin{align}
L_n^2L_{n+1} &= L_n^2(L_n + L_{n-1}) \\
&= L_n^3 + L_nL_{n-1}L_n \\
&= L_n^3 + L_nL_{n-1}(L_{n-1} + L_{n-2}) \\
&= L_n^3 + L_{n-1}^2L_n + L_nL_{n-1}L_{n-2},
\end{align}
and thus with the induction hypothesis
\begin{align}
\sum_{k = 0}^n L_k^3 &= \Biggl(\sum_{k = 0}^{n-1} L_k^3\Biggr) + L_n^3 \\
&= \Biggl(L_{n-1}^2 L_n - \sum_{m = 1}^{n-3} L_mL_{m+1}L_{m+2}\Biggr) + L_n^3 \tag{I.H.} \\
&= \Biggl(L_{n-1}^2 L_n - \sum_{m = 1}^{n-3} L_mL_{m+1}L_{m+2}\Biggr) + \bigl(L_n^2L_{n+1} - L_{n-1}^2L_n - L_{n-2}L_{n-1}L_n\bigr) \tag{see above} \\
&= L_n^2L_{n+1} - \sum_{m = 1}^{n-2} L_mL_{m+1}L_{m+2}.
\end{align}
</s>