<s>[INST] 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.
 [/INST] $\dfrac{(1+x)}{(1-x)^3}^3 =\dfrac{6}{(1-x)} - \dfrac{12}{(1-x)^2} + \dfrac{8}{(1-x)^3}  -1=6\sum_{n=0}^\infty x^n-12\sum_{n=0}^\infty (n+1)x^n+4\sum_{n=0}^\infty (n+1)(n+2)x^n=1+\sum_{n=1}^\infty (6-12n-12+4n^2+12n+8)x^n=1 + \displaystyle\sum_{n=1}^{\infty} (4n^2+2)x^n$
</s>