Answer
$$\dfrac{x^3}{1-x} ; |x| \lt 1 $$
Work Step by Step
The Taylor series for $\dfrac{1}{1-x} $ can be defined as: $\dfrac{1}{1-x}= 1+x+x^2+......+x^n+.....; |x| \lt 1$
Consider the given series:
$\\=x^3+x^4+x^5+x^6+x^7+.....\\=x^3(1+1+1+1+1....)\\=x^3 (\dfrac{1}{1-x})\\=\dfrac{x^3}{1-x} ; |x| \lt 1 $
