Find a coefficient of $x^{12}$ in $(1+x^{3} +x^{6} +x^{9} +...)^{7}$

Question

Find a coefficient of $x^{12}$ in $(1+x^{3} +x^{6} +x^{9} +...)^{7}$

I don't really have idea how to solve that. I have only realized that we can get $12$ by: $$ 12 = 12 + 0$$ $$ 12 = 9 + 3$$ $$ 12 = 6 + 6$$

But I can't see any further step that should be taken

Answer 1

I presume that $1+x^3+x^6+x^9+\cdots$ is really $\sum_{n=0}^\infty x^{3n}$ which equals $1/(1-x^3)$. Its seventh power is $1/(1-x^3)^7$. This can be expanded by the binomial theorem as $$(1-x^3)^{-7}=\sum_{n=0}^\infty {n+6\choose 6}x^{3n}$$ etc.