Let $F_n, L_n$ be the Fibonacci and Lucas sequences respectively. Show that $F_{3n} = F_{n}(L_{2n} + (-1)^n)$. In my attempt I am using Binet's formula, and the equivalent for the Lucas numbers.
\begin{align} F_{3n} &= \frac {\alpha^{3n} - \beta^{3n}} { \sqrt{5}} = F_{n}(L_{2n} + (-1)^n) \\ &= \frac {\alpha^{n} - \beta^{n}} {\sqrt{5}}(\alpha^{2n}+\beta^{2n}+(-1)^n) \\ &=\frac {\alpha^{3n} + \alpha^n\beta^{2n}+(-1)^n\alpha^n-\beta^n\alpha^{2n}-\beta^{3n}-(-1)^n\beta^n}{\sqrt{5}}, \end{align} from here I am not sure how to proceed. Hints appreciated.
Starting with $F_{3n}$ then \begin{align} F_{3n} &= \frac{1}{\sqrt{5}} \, ( \alpha^{3n} - \beta^{3n} ) \\ &= \frac{1}{\sqrt{5}} \, (\alpha^{3n} - 3 \, \alpha^{2n} \beta^n + 3 \, \alpha^n \beta^{2n} - \beta^{3n}) + \frac{3 (-1)^n}{\sqrt{5}} \, (\alpha^n - \beta^n) \\ &= \frac{1}{\sqrt{5}} \, (\alpha^n - \beta^n)^3 + 3 \, (-1)^n \, F_{n} \\ &= F_{n} \, (5 F_{n}^2 + 3(-1)^n) \\ &= F_{n} \, ( (\alpha^n - \beta^n)^2 + 3 (-1)^n) = F_{n} \, (\alpha^{2n} + \beta^{2n} + (-1)^n) \\ &= F_{n} \, (L_{2n} + (-1)^n). \end{align}