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

Hints :


*

*First, prove that 
$$\frac{1}{(1-x^2)(1-x^3)(1-x^4)} = \frac{7}{32(x+1)}-\frac{59}{288(x-1)}+\frac{1}{8(x-1)^2}+\frac{1}{16(x+1)^2}-\frac{1}{24(x-1)^3}+\frac{x+2}{9(x^2+x+1)}+\frac{1-x}{8(x^2+1)}.$$

*Then, use that
$$ \frac{1-x}{x^2+1} = -\frac{1+i}{2(x-i)}+\frac{-1+i}{2(x+i)}.$$
and
$$ \frac{x+2}{x^2+x+1} = \frac{j}{x+j}+\frac{\overline{j}}{x+\overline{j}}$$ where $j = \frac{1+i\sqrt{3}}{2}$.

*Finally, use the classical series
$$ \sum_{n=0}^{+\infty} x^k = \frac{1}{1-x},$$
$$ \sum_{n=0}^{+\infty} (k+1)x^k = \frac{1}{(1-x)^2},$$
and
$$ \sum_{n=0}^{+\infty} \frac{1}{2}(k+1)(k+2)x^k = \frac{1}{(1-x)^3}.$$

