Solve for all $x$ such that $x^3 = 2x + 1, x^4 = 3x + 2, x^5 = 5x + 3, x^6 = 8x +5 \cdots$ Question:
Solve for all $x$ such that $\begin{cases}&{x}^{3}=2x+1\\&{x}^{4}=3x+2\\&{x}^{5}=5x+3\\&{x}^{6}=8x+5\\&\vdots\end{cases}$.

My attempt:
I sum up everything.
$$\begin{aligned}\sum_{i=1}^n x^{i+2} &= (2 + 3 + 5 + 8 + \cdots)x + (1 + 2 + 3 + 5 +...)\\&= (S_n - 2)x + (S_n - 1)\\& = (a_{n+2} - 3)x + (a_{n+2} - 2)\end{aligned}$$
where $S_n$ is sum of first $n$ terms of Fibonacci series and $a_n$ is its $n$th term.
I'm not getting any idea how to solve it further.

Hint:
As $x(2x+1)\ne0,$
$$\dfrac{x^4}{x^3}=\dfrac{3x+2}{2x+1}\iff x^2-x-1=0$$
Similarly check, $$\dfrac{x^5}{x^4}=?,\dfrac{x^6}{x^5}=?$$
Find the common root of all?
