<s>[INST] Discrepancy in finding the coefficient of a polynomial After asking a question here let us consider on the following case which is simpler, this might help me to make some progress: 
suppose we have the following polynomial: 
$$f(x)=(x^0+x^1+x^2+x^3)^3$$ expanding this gives: 
$$f(x)=1 + 3 x + 6 x^2 + 10 x^3 + 12 x^4 + 12 x^5 + 10 x^6 + 6 x^7 + 
 3 x^8 + x^9$$
Aim is to find the coefficients for a given power. Now let us use multinomial theorem. Let us now call: 
$$a=x^0$$
$$b=x^1$$
$$c=x^2$$
$$d=x^3$$
then the polynomial becomes: 
$$f(a,b,c,d)=(a+b+c+d)^3$$
which has the following expansion: 
$$f(a,b,c,d)=a^3 + 3 a^2 b + 3 a b^2 + b^3 + 3 a^2 c + 6 a b c + 3 b^2 c + 
 3 a c^2 + 3 b c^2 + c^3 + 3 a^2 d + 6 a b d + 3 b^2 d + 6 a c d + 
 6 b c d + 3 c^2 d + 3 a d^2 + 3 b d^2 + 3 c d^2 + d^3$$
surely upon substituting we recover the original polynomial or its expansion. Now as a matter of example say we want to determine the coefficient of $a^2d$, we can see that the coefficient is 3, we can derive this as: 
$$a^2d: \frac{3!}{2!1!}=3$$
which is the write coefficient, now we have 
$$3a^2d=3(x^0)^2(x^3)=3x^3$$
which is not the coefficient of the expanded original polynomial, the expanded original polynomial has coefficient equal to 10, now question is what am I missing here? why there is discrepancy in here?
 [/INST] $$f(x)=\frac{(1-x^4)^3}{(1-x)^3}$$
We have that $$\frac{1}{(1-x)^3}=\sum_{k=0}^{\infty}\binom{k+2}{2}x^k.$$
And $(1-x^4)^3=1-3x^4+3x^8-x^{12}.$
So $f(x)=\sum_{k=0}^{\infty} a_kx^k$ where:
$$a_k = \binom{k+2}{2}-3\binom{k-2}{2}+3\binom{k-6}{2}-\binom{k-10}{2}.$$
More general, if $$\begin{align}f_{n,m}(x)&=\left(1+x+x^2+\cdots+x^n\right)^m\\&=\frac{\left(1-x^{n+1}\right)^m}{(1-x)^m}\\&=\left(1-x^{n+1}\right)^m\sum_{k=0}^{\infty}\binom{k+m-1}{m-1}x^k
\end{align}$$
From this we get that $$a_k=\sum_{j=0}^{m}\left(-1\right)^j\binom{m}{j}\binom{k+m-1-j(n+1)}{m-1}$$
When $n=m,$ this gives:
$$a_k=\sum_{j=0}^{n}\left(-1\right)^j\binom{n}{j}\binom{k-(n+1)(j-1)-2}{n-1}$$
There is no simpler form, I strongly suspect.
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