Chapter 10: Infinite Sequences and Series - Section 10.10 - The Binomial Series and Applications of Taylor Series - Exercises 10.10 - Page 633: 49

$$\dfrac{x^3}{1+x^2}; |x| \lt 1$$

Recall the Taylor series for $\dfrac{1}{ (1-x)}=1+x+x^2+.....x^n+....; |x| \lt 1$ Now, $=x^3-x^5+x^7-x^9-...$ or, $=x^3(1-x^2+x^4-x^6+....)$ or, $=x^3(1-x^2+(x^2)^2-(x^2)^3+.......) $ or, $ = x^3 (\dfrac{1}{1+x^2})$ or, $=\dfrac{x^3}{1+x^2}; |x| \lt 1$