Rewrite before choosing ascending powers

Both copies print the complete working. Write $(8+x)^{1/3}=[8(1+x/8)]^{1/3}=8^{1/3}(1+x/8)^{1/3}$. The source callout rewrites $(a+bx)^n$ as $a^n(1+bx/a)^n$ when the bracket does not start with 1. Here the positive factor 8 and real cube root cause no branch issue. Independently, the rewriting requires $a\ne0$; arbitrary real fractional powers also need an appropriate real domain.

$\begin{aligned}&(8+x)^{1/3}\\ & =2\left[1+\frac13\frac x8+\frac{(1/3)(1/3-1)}{2!}\left(\frac x8\right)^2+\cdots\right].\end{aligned}$

The full printed simplification is

$2\left[1+\frac{x}{24}+\frac{(1/3)(-2/3)}2\frac{x^2}{64}+\cdots\right]$

$\begin{aligned}&=2\left(1+\frac{x}{24}-\frac{x^2}{576}+\cdots\right)\\ & =2+\frac{x}{12}-\frac{x^2}{288}+\cdots.\end{aligned}$

The source gives $|x/8|<1\Rightarrow |x|<8\Rightarrow -8<x<8$. Its second callout says to replace the standard argument by $bx/a$ and hence use $|bx/a|<1$. This is the printed sufficient interval; it is not a claim that endpoint convergence has been analysed. A finite quadratic approximation needs a separate accuracy assessment.

The source first tries replacing the standard argument by $-1/x$:

$\begin{aligned}&\left(1-\frac1x\right)^{-1}\\ & =1+(-1)\left(-\frac1x\right)\\ &{}+\frac{(-1)(-2)}{2!}\left(-\frac1x\right)^2\end{aligned}$

$\hspace{1em}+\frac{(-1)(-2)(-3)}{3!}\left(-\frac1x\right)^3+\frac{(-1)(-2)(-3)(-4)}{4!}\left(-\frac1x\right)^4+\cdots$

$=1+x^{-1}+x^{-2}+x^{-3}+x^{-4}+\cdots.$

These are descending powers of x, whereas the question asks for ascending powers. Independently, this geometric series is valid for $|x|>1$; it is a different representation on a different region, not an algebraically meaningless series.

The source instead uses

$\begin{aligned}&\left(1-\frac1x\right)^{-1}\\ & =\left(\frac1x\right)^{-1}(x-1)^{-1}\\ & =x(-1)^{-1}(1-x)^{-1}\\ & =-x(1-x)^{-1}.\end{aligned}$

Its binomial substitution is

$(-x)\left[1+(-1)(-x)+\frac{(-1)(-2)}{2!}(-x)^2+\frac{(-1)(-2)(-3)}{3!}(-x)^3+\cdots\right].$

Hence the required terms are $-x-x^2-x^3-x^4+\cdots$. Keep the bracket through $x^3$ because multiplication by $-x$ supplies the fourth power.

The source gives $|-x|<1\Rightarrow |x|<1$, followed by $-1<x<1$, $x\ne0$. Its callout explains that $1/x$ is undefined at zero. Retain that exclusion even though $-x/(1-x)$ has a value there. Independently, the simplified function is an analytic extension at zero; the series represents the original expression only on the punctured interval $(-1,0)\cup(0,1)$.