Alternatively, when you see the alternating signs, split it into two interlocking subsequences, $81,167,295,473,709,1011,\dots$ and $123,229,381,587,855,1193,\dots$. The first subsequence has first differences $86,128,178,236,302,\dots$, second differences $42,50,58,66,\dots$, and third differences $8,8,8,\dots$, and the second subsequence has first differences $106,152,206,268,338,\dots$, second differences $46,54,62,70,\dots$, and third differences $8,8,8,\dots$. Work with these separately.
If the first subsequence is $x_0,x_1,\dots$, its first differences are $\Delta x_0,\Delta x_1,\dots$, and its second differences are $\Delta^{(2)} x_0,\Delta^{(2)} x_1,\dots$. The third differences are apparently constant at $8$, so it should be clear that $\Delta^{(2)} x_n = 42 + 8n$. This means that $\begin{align*} \Delta x_n &= 86 + \sum\limits_{k=0}^{n-1} (42 + 8k)\\ &= 86 + 42n + 8\sum\limits_{k=0}^{n-1}k\\ &= 86 + 42n + 8 \left(\frac{n(n-1)}{2}\right)\\ &= 86 + 42n + 4n(n-1)\\ &= 86 + 38n + 4n^2. \end{align*}$ Repeat the process one more time: $\begin{align*} x_n &= 81 + \sum\limits_{k=0}^{n-1}(86 + 38k + 4k^2)\\ &= 81 + 86n + 38\sum\limits_{k=0}^{n-1}k + 4\sum\limits_{k=0}^{n-1}k^2\\ &= 81 + 86n + 19n(n-1) + 4\left(\frac{(n-1)n(2n-1)}{6}\right)\\ &= 81 + 67n + 19n^2 + \frac{2}{3}n(2n^2-3n+1)\\ &= 81 + \frac{203}{3}n + 17n^2 + \frac{4}{3}n^3. \end{align*}$
Exactly the same sort of analysis can then be applied to the second subsequence. I’ll leave you to check it, but if I’m not mistaken, you should get $y_n = 123+\frac{257}{3}n+19n^2+\frac{4}{3}n^3$.
It’s entirely possible to sew these two formulas together to get a single formula for the original sequence. If the original sequence is $a_0,a_1,\dots$, then $x_n = a_{2n}$, and $y_n = a_{2n+1}$. Thus, $a_{2n} = 81 + \frac{203}{3}n + 17n^2 + \frac{4}{3}n^3,$ and $a_{2n+1} = 123+\frac{257}{3}n+19n^2+\frac{4}{3}n^3.$ These can be rewritten as $a_n = 81 + \frac{203}{3}\left\lfloor \frac{n}{2} \right\rfloor + 17 \left\lfloor \frac{n}{2} \right\rfloor^2 + \frac{4}{3} \left\lfloor \frac{n}{2} \right\rfloor^3$ when $n$ is even, and $a_n =$ $123 + \frac{257}{3} \left\lfloor \frac{n}{2} \right\rfloor + 19 \left\lfloor \frac{n}{2} \right\rfloor^2 + \frac{4}{3} \left\lfloor \frac{n}{2} \right\rfloor^3$ when $n$ is odd. Note that $18-(-1)^n$ is $17$ when $n$ is even and $19$ when $n$ is odd. Similarly, $230-27(-1)^n$ is $203$ when $n$ is even and $257$ when $n$ is odd, and $102-21(-1)^n$ is $81$ when $n$ is even and $123$ when $n$ is odd. Thus, the two expressions can be combined to yield $a_n = 102-21(-1)^n + \frac{230-27(-1)^n}{3}\left\lfloor \frac{n}{2} \right\rfloor + (18-(-1)^n)\left\lfloor \frac{n}{2} \right\rfloor^2 + \frac{4}{3}\left\lfloor \frac{n}{2} \right\rfloor^3.$ Finally $\left\lfloor \frac{n}{2} \right\rfloor = \frac{n}{2}+\frac{(-1)^n-1}{4} = \frac{2n-1+(-1)^n}{4}$; you can substitute this into the expression for $a_n$ and simplify to get an expression of the form $a_n = p(n)+(-1)^nq(n)$ for some polynomials $p$ and $q$.
This is certainly not the most efficient way to go about solving the problem, but it does show how you could have reduced the problem to two problems of a more familiar type.