Closed form of recurrence relation $a_{n}=a_{n-1}(a_{n-2}^{2}-a_{0})-a_{1}$ Given the following sequence: 

$$\begin{cases}a_{0}=2\\a_{1}=\frac{5}{2}\\a_{n}=a_{n-1}(a_{n-2}^{2}-a_{0})-a_{1}\end{cases}$$

Prove that
$$a_{n}=2^{\frac{2^{n}-(-1)^{n}}{3}}+2^{\frac{(-1)^{n}-2^{n}}{3}}$$
For all natural numbers $n$.

My try: (using induction)
Let $b_{n}=\frac{2^{n}-(-1)^{n}}{3}$
For $n=0$: correct $\checkmark$ 
Now for $n+1$:
$$
\begin{aligned}
a_{n+1} &= a_{n}(a_{n-1}^{2}-a_{0})-a_{1})=(2^{b_{n}}+2^{-b_{n}})(2^{2b_{n}}+2^{-2b_{n}}+2) \\
&=2^{3b_{n}}+2^{-3b_{n}}+2^{b_{n}}+2^{-b_{n}}+2^{b_{n}+1}+2^{-b_{n}+1}
\end{aligned}
$$
How do I proceed from here?

You want to proof that $a_n = f(n)$ for any $n\inℕ$ with
$$ f(n) := 2^{\frac{2^{n}-(-1)^{n}}{3}}+2^{\frac{(-1)^{n}-2^{n}}{3}}.$$
Check of two (!) initial conditions:


*

*$f(0) = 2 = a_0 \checkmark$

*$f(1) = \frac52 = a_1 \checkmark$
Inductive step:
$$\begin{aligned}
a_{n+1} &= a_n (a_{n-1}^2 - a_0) - a_1\\
&= f(n) (f(n-1)^2 - f(0)) - f(1)\\
&= (2^{b_n} + 2^{-b_n}) \cdot ((2^{b_{n-1}} + 2^{-b_{n-1}})^2 - 2) - \frac52\\
&= (2^{b_n} + 2^{-b_n}) \cdot (2^{2b_{n-1}} + 2^{-2b_{n-1}}) - \frac52 \\
&= 2^{b_n+2b_{n-1}} + 2^{b_n-2b_{n-1}} + 2^{-b_n+2b_{n-1}} +  2^{-b_n-2b_{n-1}} - \frac52 \end{aligned}$$
Simple computations show that:
$$\begin{aligned}
b_n+2b_{n-1} &= \frac13(2^{n+1} + (-1)^n) = b_{n+1}\\
b_n-2b_{n-1} &= (-1)^{n+1}.
\end{aligned}$$
Thus
$$a_{n+1} = f(n+1) + 2^{(-1)^{n+1}} + 2^{(-1)^n} - \frac52$$
With the equality (for $n\in\mathbb N$)
$$2^{(-1)^{n+1}} + 2^{(-1)^n} = \frac52$$
we can finish the proof:
$$a_{n+1} = f(n+1).$$
