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.

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.
