I'm trying to find the 15th term's coefficient in the following equation.
$$\frac{\left(1-x^6\right)^4}{(1-x)^4}$$
What I do is segregate each term as follows respectively.
$$\left(1-x^6\right)^4 \\ \text{and} \\ \frac{1}{(1-x)^4}$$
I think the multiplication of coefficients of $x^{12}$ and $x^{3}$ is the adequate way to obtain the coefficient of $x^{15}$.
$$\left(1-x^6\right)^4 = x^{24}-4 x^{18}\color{blue}{+6 x^{12}}-4 x^6+1$$
$$\frac{1}{(1-x)^4} = 1+4 x+10 x^2\color{blue}{+20 x^3}+35 x^4+56 x^5+... $$
It yields $\color{blue}{120 x^{15}}$. However, I get the following when I put the first composite expression to obtain its Taylor series.
$$\frac{\left(1-x^6\right)^4}{(1-x)^4} = \frac{O}{x^6}+x^{20}+4 x^{19}+10 x^{18}+20 x^{17}+35 x^{16}\color{purple}{+56 x^{15}}+80 x^{14}+104 x^{13}+125 x^{12}+140 x^{11}+146 x^{10}+140 x^9+125 x^8+104 x^7+80 x^6+56 x^5+35 x^4+20 x^3+10 x^2+4 x+1$$
Where did I go wrong?
There are a couple of problems: the first one is that the last line you wrote represents the expansion around $\infty$ and not around $0$. Moreover, to get the coefficient of $x^{15}$, you have to consider the coefficients of: $x^{12}$ and $x^3$, $x^6$ and $x^9$, $x^{15}$ and $x^0$. The correct answer is the following: $$(1-x^6)^4 = x^{24}-4 x^{18}+\color{blue}6 x^{12}-\color{orange}4 x^6+\color{red}1\\ \frac1{(1-x)^4}=1+4x+\dots+\color{blue}{20}x^3+\dots+\color{orange}{220}x^9+\dots+\color{red}{816}x^{15}+\dots $$ The coefficient of $x^{15}$ is: $$6\cdot20+ 220\cdot(-4)+816\cdot 1=56$$ and it matches with the expansion around $x=0$.