Practise finite and nonterminating binomial expansions

Both copies give the answer without working. Independently, the finite binomial theorem gives

$\begin{aligned}&(4+x)^4\\ & =4^4+4(4^3)x+6(4^2)x^2\\ &{}+4(4)x^3+x^4.\end{aligned}$

The printed answer is $256+256x+96x^2+16x^3+x^4$, for $x\in\mathbb R$. It includes the fourth-degree term although the common prompt asks only through $x^3$. The requested truncation is $256+256x+96x^2+16x^3$; the full finite identity above preserves the supplied answer. The interval is all real x; the cubic truncation itself is an approximation with exact omitted term $x^4$.

Independently rewrite $(1/x+1)^{-2}=x^2(1+x)^{-2}$ for $x\ne0,-1$. Then

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

This matches the printed answer. The expansion condition is $|x|<1$, with the original exclusion $x\ne0$, so the printed set is $\{x\in\mathbb R:-1<x<1,\ x\ne0\}$. Terms through the bracket’s first power suffice for the requested cubic result.

Independently write $(x+4)^{-3/2}=\tfrac18(1+x/4)^{-3/2}$ and use the falling-factorial coefficients:

$\frac18\left[1-\frac32\frac x4+\frac{(-3/2)(-5/2)}{2!}\left(\frac x4\right)^2+\frac{(-3/2)(-5/2)(-7/2)}{3!}\left(\frac x4\right)^3+\cdots\right].$

The printed answer is

$\frac18\left(1-\frac{3x}{8}+\frac{15x^2}{128}-\frac{35x^3}{1024}+\cdots\right),\qquad \{x\in\mathbb R:-4<x<4\}.$

Transform $|x/4|<1$ to obtain that interval. The original real function’s domain $x>-4$ is larger than this series interval.