If you are willing to deal with complex numbers, you can use a partial fraction expansion.
For any polynomial $\displaystyle P(x) = \prod_{j=1}^{n}(x-\alpha_j)$ with distinct roots, we have that
\frac{1}{P(x)} = \sum_{j=1}^{n} \frac{1}{P'(\alpha_j)(x-\alpha_j)} = -\sum_{j=1}^{n} \frac{1}{\alpha_jP'(\alpha_j)(1- \frac{x}{\alpha_j})}
You can expand out every $\displaystyle \frac{1}{1 - \frac{x}{\alpha_j}}$ in an infinite power series and add them to give on single series, giving a formula for the coefficients, in terms of the $\displaystyle \alpha_j$.
In your case, we can write the denominator as $\displaystyle (1-x)^3 P(x)$, where $\displaystyle P(x)$ is a polynomial with distinct complex roots, and the above partial fraction expansion can be used as
\frac{1}{(1-x)^3 P(x)} = \frac{1}{(1-x)^3}\sum_{j=1}^{n} \frac{1}{P'(\alpha_j)(x-\alpha_j)}
Now you can use the binomial theorem (yes, it works for negative numbers, but you get an infinite series) for the $\displaystyle \frac{1}{(1-x)^3}$ portion and you get a closed form formula, involving complex numbers, which you can ultimately convert into a "real number only" formula using $\displaystyle \cos$ and $\displaystyle \sin$.
Or if you want to simplify it further, you can try using an expansion of the form
$\frac{1}{(1-x)^3 P(x)} = \frac{A}{1-x} + \frac{B}{(1-x)^2} + \frac{C}{(1-x)^3} + \sum_{j=1}^{n} \frac{D_j}{x-\alpha_j}$
which should give a closed form, which uses a "constant number of terms" to compute the coefficient of $x^N$.