Finding coefficient of polynomial? The coefficient of $x^{12}$ in $(x^3 + x^4 + x^5 + x^6 + …)^3$ is_______?

My Try:
Somewhere it explain as:
The expression can be re-written as: $(x^3 (1+ x + x^2 + x^3 + …))^3=x^9(1+(x+x^2+x^3))^3$
Expanding $(1+(x+x^2+x^3))^3$ using binomial expansion:
$(1+(x+x^2+x^3))^3 $
$= 1+3(x+x^2+x^3)+3*2/2((x+x^2+x^3)^2+3*2*1/6(x+x^2+x^3)^3…..$
The coefficient of $x^3$ will be $10$, it is multiplied by $x^9$ outside, so coefficient of $x^{12}$ is $10$.

Can you please explain?


Another way: For $|x|<1$, we have:
$$(x^3+x^4+x^5+...)^3=x^9(1+x+x^2+...)^3=x^9(1-x)^{-1}.$$
Now $(1-x)^{-3}$ is  half of the second derivative of $(1-x)^{-1}.$ The second derivative of  $(1-x)^{-1}=(1+x+x^2+...)$ is $(1.2+2.3 x+3.4 x^2+4.5 x^3+...). $ Half the co-efficient of $x^3$ in this, which is $(1/2).4.5=10,$ is therefore the co-efficient of $x^{12}$ in $x^9(1+x+x^2+...). $
