<s>[INST] if $\frac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$,find $a_{n}$ Let 
$$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$$
Find the closed form  $$a_{n}$$
since
$$(1-x^4)(1-x^3)(1-x^2)=(1-x)^3(1+x+x^2+x^3)(1+x+x^2)(1+x)$$
so
$$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\dfrac{1}{(1-x)^3(1+x+x^2+x^3)(1+x+x^2)(1+x)}$$
then I feel very ugly,can you someone have good  partial fractions methods by hand?
because I take an hour to solve this problem. 
ago I have solve $$x+2y+3z=n$$ the number of the positive integer solution $a_{n}$ I found
$$a_{n}=\left[\dfrac{(n+3)^2}{12}\right]$$
Thank you
 [/INST] Incomplete answer (maybe it can be used):
If $\left|z\right|<1$ then $\frac{1}{1-z}=1+z+z^{2}+\cdots$
Applying this for $z=x^{4},x^{3},x^{2}$ you find:
$$\frac{1}{\left(1-x^{4}\right)\left(1-x^{3}\right)\left(1-x^{2}\right)}=\left(1+x^{4}+x^{8}+\cdots\right)\left(1+x^{3}+x^{6}+\cdots\right)\left(1+x^{2}+x^{4}+\cdots\right)$$
Then $a_{n}$ must equalize the cardinality of $\left\{ \left(p,q,r\right)\in\left\{ 0,1,2,\dots\right\} \mid4p+3q+2r=n\right\} $ (as Macavity comments). 
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