It seems like that the order we choose influence the formula, but it shouldn't be. It should be like $(a+b)^{-n}=\sum_{k=0}^{\infty}\binom {-n}{k}a^k b^{-n-k}=\sum_{k=0}^{\infty}\binom {-n}{k}a^{-n-k}b^{k}$.
Here is the example why I'm confused. $$f(x):=((1-x)^3-x^3)^{-1}$$ We can get $f(x)=\sum_{k=0}^{\infty}\binom {-1}{k}((1-x)^{3})^{k} (-x^3)^{-1-k}=\sum_{k=0}^{\infty}-(1-x)^{3k} x^{-3-3k}$ As we can see, there doesn't have any term like $a^n,n\ge 0$ because $(1-x)^{3k}$ only have terms $x^0, x^1,\ldots,x^{3k}$.
But if we rewrite it, we can get every term in $f(x)$ has such form, $a^n,n\ge 0$.$f(x)=\sum_{k=0}^{\infty}\binom {-1}{k}((1-x)^{3})^{-1-k} (-x^3)^{k}=\sum_{k=0}^{\infty}(1-x)^{-3-3k} x^{3k}=\sum_{k=0}^{\infty}\sum_{i=0}^{\infty}\binom {-3-3k}{i}(-x)^{i} x^{3k}$, then all the terms are like $a^n,n\ge 0$.
There is an important assumption that is missing. When $r\in \mathbb{C}$
$$(a+b)^{r} = \begin{cases} \displaystyle \sum_{k=0}^{\infty}\binom {r}{k}a^k b^{r-k} \quad \textrm{ if } \quad |a|< |b|\\ \displaystyle \sum_{k=0}^{\infty}\binom {r}{k}a^{r-k}b^{k}\quad \textrm{ if } \quad |b|<|a|\end{cases}$$
The hypotehsis of $|a|<|b|$ is to guarantee convergence.
So, to calculate your series, supposing that $x\in \mathbb{R}$, you need to know the domain of the function ie. if
$$ |1-x|^3 > |x|^3 \Longleftrightarrow x\in \left(-\infty,\frac{1}{2}\right)$$
you should use
$$((1-x)^3-x^3)^{-1} = \sum_{i=0}^{\infty}\binom {-1}{i}(-1)^{i}(1-x)^{-3-3i}x^{3i} $$
but if
$$ |1-x|^3 < |x|^3 \Longleftrightarrow x\in \left(\frac{1}{2},\infty\right)$$
you should use
$$((1-x)^3-x^3)^{-1} = \sum_{i=0}^{\infty}\binom {-1}{i}(-1)^{-1-i}(1-x)^{3i}x^{-3-3i} $$
Sometimes the case $|a|=|b|$ also converges, but you need to check it.