Please explain this asymptotic equation

Question

From Valdimir A.Zorich, Mathematical Analysis I, pp. 144-145, \begin{align} (1+x)^{\alpha}&=1+\frac{\alpha}{1!} x+\frac{\alpha(\alpha-1)}{2!} x^2+ \dotsb+ \\ &\phantom{={}}+\frac{\alpha(\alpha-1)\dotsb(\alpha-n+1)}{n!}x^n+O(x^{n+1})\quad\text{as }x\to 0 \end{align} Example 44, as $x\to\infty$ we have \begin{align} \frac{x^3+x}{1+x^3}&=\frac{1+x^{-2}}{1+x^{-3}} = \left( 1+ \frac{1}{x^2} \right)\left(1+ \frac{1}{x^3} \right)^{-1} \\ & = \left( 1+ \frac{1}{x^2} \right) \left(1- \frac{1}{x^3} +O\left(\frac{1}{x^6}\right)\right) = \left(1+ \frac{1}{x^2} +O\left(\frac{1}{x^3}\right)\right) \end{align} I don’t understand this equation (left to right side) \begin{align} \left( 1+ \frac{1}{x^2} \right) \left(1- \frac{1}{x^3} +O\left(\frac{1}{x^6}\right)\right) = \left(1+ \frac{1}{x^2} +O\left(\frac{1}{x^3}\right)\right) \end{align} Please explain it, thanks.

Answer 1

Multiply out:

$\begin{align*} &\left( 1 + \frac{1}{x^2} \right) \left( 1 - \frac{1}{x^3} + O\left(\frac{1}{x^6}\right) \right) \\ &\qquad= \left( 1 + \frac{1}{x^2} - \frac{1}{x^3} - \frac{1}{x^5} + \frac{1}{x^2} O\left(\frac{1}{x^6}\right) \right) \\ &\qquad= \left( 1 + \frac{1}{x^2} - \frac{1}{x^3} - \frac{1}{x^5} + O\left(\frac{1}{x^8}\right) \right) \\ &\qquad= \left( 1 + \frac{1}{x^2} + O\left(\frac{1}{x^3}\right) \right) \end{align*}$

Last step throws away some precision by cutting off after the second term.