I'm trying to prove something about two linearly independent solutions; $a_n$, $b_n$, to a recurrence relation I have - specifically that $\left| \frac{a_n}{b_n} \right|$ is eventually monotonically increasing using an estimate of the linearly independent solutions. The recurrence relation in question is: \begin{align*} n^3u_n&=-3\left(27(n-1)^3-8(n-1)-4\right)u_{n-1}\\&\quad {}-3^4\left(27(n-2)^3-8(n-2)+4\right)u_{n-2}-3^9\left(n-3\right)^3u_{n-3} \end{align*}
and the first few terms of the solutions $a_n$, $b_n$ are: \begin{align*} (a_n)_{n\geq0} &= \{0, 1/12, 3/32, -133/24, \ldots\} \\ (b_n)_{n\geq0} &= \{1/12, 1, -9, 127, \ldots\} \end{align*}
Unfortunately I've only ever seen recurrence relations of order up to 2, and this isn't really my area of knowledge at all. Would anyone happen to have any pointers? I've tried to use Mathematica's RSolve function, but it didn't yield anything helpful. Any advice (though not an explicit solution) would be very much appreciated! Thanks.
If you set $t_n = n^3 u_n$, you can approximate your recurrence with a linear one: $$ t_n + 3^4 t_{n+1} + 3^7 t_{n-2} + 3^9 t_{n-3} \sim 0, \tag{1} $$ then study the roots of the characteristic polynomial: $$ p(\lambda) = \lambda^3 + 3^4 \lambda^2 + 3^7 \lambda + 3^9 = (\lambda+27)^3, \tag{2}$$ in order to have that $t_n$ behaves like $(-27)^n$, so $u_n$ behaves like $(-1)^n\cdot\frac{27^n}{n^3}.$