Suppose $ \sum_{n=1}^{\infty} b_n x^n= \frac{x^3}{(x^4-1)^2}$. What could be an expression of $b_n$? A practice problem reads:
Suppose $$\sum_{n=1}^{\infty} b_n x^n = \frac{x^3}{(x^4-1)^2}.$$
What could be an expression of $b_n$? 
Some of the possible answers read $ 2^{3n}nx^{3n-1}, nx^{3n-1}, nx^{4n-1}$
There are a few other answers. I was just not sure how to start this problem? 

Another variation is based  upon the binomial series expansion.
\begin{align*}
(1+x)^\alpha=\sum_{n=0}^\infty\binom{\alpha}{n}x^n\qquad\qquad |x|<1
\end{align*}

We obtain
  \begin{align*}
\frac{x^3}{(x^4-1)^2}&=x^3\sum_{n=0}^\infty\binom{-2}{n}(-x^4)^n\tag{1}\\
&=x^3\sum_{n=0}^\infty\binom{n+1}{1}x^{4n}\\
&=\sum_{n=0}^\infty  (n+1)x^{4n+3}\tag{2}\\
&=x^3 + 2 x^7 + 3 x^{11} + 4 x^{15} + 5 x^{19} + 6 x^{23}+\cdots
\end{align*}

In (1) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

We conclude according to (2) the representation
  \begin{align*}
\sum_{n=1}^{\infty} b_n x^n=\sum_{n=0}^\infty  (n+1)x^{4n+3}
\end{align*}
  implies for $n\geq 0$
  \begin{align*}
b_{4n+k}=
\begin{cases}
n+1&\qquad k=3\\
0&\qquad k\neq 3
\end{cases}
\end{align*}

