Finding the coefficient of $x^r$ in an expansion. Suppose that the summation of the infinite series $$1+nx+\frac{n(n-1)}{2} x^2+\cdots+\frac{n(n-1)\cdots(n-r+1)}{r}x^r+\cdots$$ is equal to $(1+x)^n$ for $|x|<1$.
Show that the coefficient of $x^r$ in the expansion of $\frac{1+x+x^2}{(1-x)^2}$  is $3r $. 
Hence show that $(217)^\frac{1}{3} \simeq 6.0092$
My attempt :
$$\frac{1+x+x^2}{(1-x)^2}=\frac{(1+x+x^2)(1-x)}{(1-x)^3}$$
$$\frac{1+x+x^2}{(1-x)^2}=\frac{1-x^3}{(1-x)^3}$$
$$=\frac{1}{(1-x)^3}-\frac{x^3}{(1-x)^3}$$
$$=\frac{1}{\left(1+(-x)\right)^3}+\frac{1}{\left(1+\left(-\frac{1}{x}\right)\right)^3}$$
$$=(1+(-x))^{-3}+\left(1+\left(-\frac{1}{x}\right)\right)^{-3}$$
How can I proceed after this ? Is there another method ? Is my method correct ? 

If we MUST use the binomial coefficients, this becomes rather tedious:
$$\frac{1+x^2+x}{(1-x)^2}=\frac{(1-x)^2+3x}{(1-x)^2}=1+\frac {3x}{(1-x)^2}$$$$=1-3x(1-x)^{-2}=1-3(1-x)^{-1}+3(1-x)^{-2}$$
$$=1-3\sum\binom{-1}{r}(-1)^rx^r+3\sum\binom{-2}{r}(-1)^rx^r$$
$$=1+3\sum \left(-\binom{-1}{r}+\binom{-2}{r}\right)(-1)^rx^r$$
Let's try to prove that the binomial difference if equal to $(-1)^r\times r$. For $r=0$ the sum is $0$, for $k=1$ the sum is $-1$, assuming this is true for $r=m$, let's check for $r=m+1$: $$-\binom{-1}{m+1}+\binom{-2}{m+1}=-\frac{-1-m}{1+m}\binom{-1}{m}+\frac{-2-m}{1+m}\binom{-2}{m}$$ $$=(-1)m(-1)^m-\frac 1{m+1}\binom{-2}{m}=m(-1)^{m+1}+(-1)^{m+1}$$
So the expansion becomes $$1+\sum_r3rx^r$$
