How can we show that $2^3+1^3+3^3+4^3+\cdots+L_n^3={L_nL_{n+1}^2+5(-1)^n[L_{n-1}+2(-1)^n]+9\over 2}$ In my recent post I was intended to asked this question
$...,3,-1,2,1,3,4,7,...$ for $n=...,-2,-1,0,1,2,3,4,...$; It is the Lucas numbers
Let the sum of the cube of Lucas series be $S_L$
$S_L=2^3+1^3+3^3+4^3+\cdots+L_n^3$,
show that it has a closed form $$S_L={L_nL_{n+1}^2+5(-1)^n[L_{n-1}+2(-1)^n]+9\over 2}$$
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,
$L_{n+1}^3+L_{n+2}^3={L_{n+3}\over 2}(L_n^2+L_{n+1}^2+L_{n+2}^2)$
$2^2+1^2+3^2+\cdots+L_n^2=L_nL_{n+1}+2$
Any further hints?
 A: This answer uses that $$L_n^2-L_{n-1}L_{n+1}=5(-1)^n\tag1$$
whose proof can be seen at the end of this answer.
Using $(1)$, we get
$$\begin{align}&{L_nL_{n+1}^2+5(-1)^n[L_{n-1}+2(-1)^n]+9\over 2}\\\\&=\frac{L_nL_{n+1}^2+5(-1)^nL_{n-1}+10(-1)^{2n}+9}{2}\\\\&=\frac{L_nL_{n+1}^2+(L_n^2-L_{n-1}L_{n+1})L_{n-1}+10+9}{2}\\\\&=\frac{L_nL_{n+1}^2-L_{n-1}^2L_{n+1}+L_{n-1}L_n^2+19}{2}\end{align}$$
So, this answer proves by induction that
$$\sum_{k=0}^{n}L_k^3=\frac{L_nL_{n+1}^2-L_{n-1}^2L_{n+1}+L_{n-1}L_n^2+19}{2}$$
It holds for $n=0$.
Supposing that it holds for some $n\ (\ge 0)$ gives that
$$\begin{align}\sum_{k=0}^{n+1}L_k^3&=L_{n+1}^3+\sum_{k=0}^{n}L_k^3\\\\&=L_{n+1}^3+\frac{L_nL_{n+1}^2-L_{n-1}^2L_{n+1}+L_{n-1}L_n^2+19}{2}\\\\&=L_{n+1}^3+\frac{L_nL_{n+1}^2-(L_{n+1}-L_n)^2L_{n+1}+(L_{n+1}-L_n)L_n^2+19}{2}\\\\&=\frac{L_{n+1}^3+2L_nL_{n+1}^2+L_n^2L_{n+1}-L_n^3-L_n^2L_{n+1}+L_nL_{n+1}^2+19}{2}\\\\&=\frac{L_{n+1}(L_{n+1}+L_n)^2-L_n^2(L_n+L_{n+1})+L_nL_{n+1}^2+19}{2}\\\\&=\frac{L_{n+1}L_{n+2}^2-L_n^2L_{n+2}+L_nL_{n+1}^2+19}{2}\qquad\blacksquare\end{align}$$

Let us prove $(1)$ by induction.
It holds for $n=0$.
Supposing that it holds for some $n\ (\ge 0)$ gives that
$$\begin{align}5(-1)^{n+1}&=5(-1)^n\cdot (-1)\\\\&=-(L_n^2-L_{n-1}L_{n+1})\\\\&=-L_n^2+L_{n-1}(L_{n-1}+L_n)\\\\&=-L_n^2+L_{n-1}^2-L_nL_{n-1}+2L_nL_{n-1}\\\\&=L_{n-1}^2+2L_nL_{n-1}-L_n(L_n+L_{n-1})\\\\&=L_{n-1}^2+2L_nL_{n-1}-L_nL_{n+1}\\\\&=L_{n-1}^2+2L_nL_{n-1}+L_n^2-L_n^2-L_nL_{n+1}\\\\&=(L_{n-1}+L_n)^2-L_n(L_n+L_{n+1})\\\\&=L_{n+1}^2-L_nL_{n+2}\qquad\blacksquare\end{align}$$
A: Some relations about Fibonacci and Lucas numbers:

