Power series for the rational function $(1+x)^3/(1-x)^3$ 
Show that $$\dfrac{(1+x)^3}{(1-x)^3} =1 + \displaystyle\sum_{n=1}^{\infty} (4n^2+2)x^n$$

I tried with the partial frationaising the expression that gives me 
$\dfrac{-6}{(x-1)} - \dfrac{12}{(x-1)^2} - \dfrac{8}{(x-1)^3}  -1$ 
how to proceed further on this having doubt with square and third power term in denominator.

As shown in this answer, $\binom{-n\vphantom{1}}{k}=(-1)^k\binom{n+k-1}{k}$, therefore,
$$
\begin{align}
\frac{(1+x)^3}{(1-x)^3}
&=(1+x)^3\sum_{k=0}^\infty\binom{-3}{k}(-x)^k\\
&=(1+x)^3\sum_{k=0}^\infty\binom{k+2}{k}x^k\\
&=\sum_{k=0}^\infty\sum_{j=0}^3\binom{3}{j}\binom{k-j+2}{k-j}x^k\\
&=1+\sum_{k=1}^\infty(4k^2+2)x^k
\end{align}
$$
since, for $k\ge1$,
$$
\begin{align}
&\overbrace{\binom{k+2}{k}}^{j=0}+\overbrace{3\binom{k+1}{k-1}}^{j=1}+\overbrace{3\binom{k}{k-2}}^{j=2}+\overbrace{\color{#C00000}{\binom{k-1}{k-3}}}^{j=3}\\[3pt]
&=\binom{k+2}{2}+3\binom{k+1}{2}+3\binom{k}{2}+\color{#C00000}{\binom{k-1}{2}}\\[3pt]
&=\frac{k^2+3k+2}2+\frac{3k^2+3k}2+\frac{3k^2-3k}2+\frac{k^2-3k+2}2\\[9pt]
&=4k^2+2
\end{align}
$$
Note that the red coefficients do not match for $k=0$; $\binom{-1}{-3}=0$ while $\binom{-1}{2}=1$. The rest match for $k=0$ and all match for $k\ge1$ because $\binom{n\vphantom{1}}{k}=\binom{n\vphantom{1}}{n-k}$ for $n\ge0$.
