<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] The most simple way to prove your identity, IMHO, is to multiply both sides by $(1-x)^3$. 
This leads to:
$$ 1+3x+3x^2+x^3\stackrel{?}{=}(1-3x+3x^2+x^3)\left(1+\sum_{n\geq 1}(4n^2+2)\,x^n\right).\tag{1}$$
If we set $a_n=(4n^2+2)$, for any $n\geq 4$ the coefficient of $x^n$ in the RHS is given by $a_n-3a_{n-1}+3a_{n-2}+a_{n-3}$ that is zero, since we are applying three times the backward difference operator to a polynomial in $n$ having degree two. So we just have to check that the first four coefficients, $[x^0],[x^1],[x^2],[x^3]$, match.
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