Power Series Coefficients Find the sum of the coefficients of $x^{20}$ and $x^{21}$ in the power series expansion of $\frac 1{(1-x^3)^4}$.

I don't know a lot on power series at the moment, and I was wondering how do I find the coefficients?
Thanks
 A: HINT:
Let $y=x^3$ and expand $f(y)=(1-y)^{-4}$ in a Taylor series around $y=0$.  
This series will have terms for $x^{3n}$ only, which implies that the coefficient for $x^{20}$ is zero.
We note that the $n$'th coefficient is $(n+3)!/3!n!$, from which we see that the coefficient for $x^{21}$ is at $n=7$ or $\binom{10}{3}$.
A: Hint:
$$\sum_{n=k}^{\infty}\binom{n}{k}y^{n-k}=\left(1-y\right)^{-k-1}$$
This can be proved by induction to $k$. 
The base case is of course $\sum_{n=0}^{\infty}y^{n}=\left(1-y\right)^{-1}$.  
The inductionstep comes to differentiating both sides and draw conclusions.
Apply this for $y=x^3$ and $k=3$
A: Recall that
$$\frac1{1-s} = \sum_{n=0}^\infty s^n. $$
Differentiating we have
$$\frac{\mathsf d^3}{\mathsf ds^3}\left[\frac1{1-s}\right] = \frac6{(1-s)^4},
$$
and
$$\begin{align*}
\frac{\mathsf d^3}{\mathsf ds^3}\left[\sum_{n=0}^\infty s^n\right] &= \sum_{n=0}^\infty \frac{\mathsf d^3}{\mathsf ds^3}[s^n]\\
&= \sum_{n=0}^\infty n(n-1)(n-2)s^{n-3}\\
&= \sum_{n=0}^\infty (n+1)(n+2)(n+3)s^n.
\end{align*}$$
Hence,
$$\frac1{(1-s)^4} = \sum_{n=0}^\infty \frac16(n+1)(n+2)(n+3)s^n. $$
Put $s=x^3$, then
$$\frac1{(1-x^3)^4} = \sum_{n=0}^\infty \frac16(n+1)(n+2)(n+3)x^{3n}. $$
It is clear that $x^{20}=0$ and
$$x^{21} = \frac16(7+1)(7+2)(7+3) =  120. $$
