Consider the following sequence $(a_n)_{n \ge 0}$ (it has positive terms):
$$a_0 = 1$$
$$a_1 = a$$
$$\hspace{3.5cm} a_{n+1}^3 = a_n^2 a_{n-1} \hspace{2cm} n \ge 1$$
I have to find the value of $a$ such that the following is true:
$$\lim\limits_{n \to \infty} a_n = 8$$
I tried finding the first few terms of the sequence in the hope that I will find some kind of pattern, but I got nowhere.
From $$a_{n+1}^3 = a_n^2 a_{n-1} \iff 3\log{a_{n+1}}=2\log{a_{n}} + \log{a_{n-1}}$$ and noting $b_n=\log{a_n}$, we have $$3b_{n+1}=2b_{n} + b_{n-1}$$ with characteristic polynomial $$3x^2-2x-1=0$$ with $x_1=-\frac{1}{3}$ and $x_2=1$ as solutions, thus the general term is $$b_n=C_1\left(-\frac{1}{3}\right)^n+C_2(1)^n=C_1\left(-\frac{1}{3}\right)^n+C_2$$ or $$a_n=e^{C_1\left(-\frac{1}{3}\right)^n+C_2}$$ $C_1,C_2$ can be found from $$\left\{\begin{matrix} 1=a_0=e^{C_1+C_2}\\ a=a_1=e^{-\frac{C_1}{3}+C_2} \end{matrix}\right.$$ or $$C_1=-C_2=-\frac{3}{4}\log{a}$$ and finally $$a_n=e^{\left(1-\left(-\frac{1}{3}\right)^n\right)\frac{3}{4}\log{a}}=a^{\left(1-\left(-\frac{1}{3}\right)^n\right)\frac{3}{4}}\to a^{\frac{3}{4}},n\to\infty$$ From $a^{\frac{3}{4}}=8 \Rightarrow a=16$.
Can you see that $a_n=a^{b_n}$ where $b_n$ is the power $a$ is raised to? We can write the given relation as $a_{n+1}=\left(a_n\right)^{2/3}\left(a_{n-1}\right)^{1/3}$ so the relation between the exponents is $b_{n+1}=\frac 23b_n+\frac 13b_{n-1}$ with initial terms $b_0=0, b_1=1$. The solution to this is $b_n=\frac 34\left((-1)^{n+1}3^{-n}+1\right)\to \frac 34 \;(n\to\infty)$ and $\lim_{n\to\infty}a_n=a^{3/4}$. Now equate to $8$.
