Answer
$x^3-x^4+\frac{x^5}{2}x-\frac{x^6}{6}+...+\frac{(-x)^{n+3}}{n!}+...$
Work Step by Step
We are given $f(x)=x^3e^{-x}$
for $r$ in $(-\infty,\infty)$
The Taylor series for $e^x$ is
$1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+...+\frac{1}{n!}x^n+...$
The Taylor series for $f(x)=x^3e^{-x}$ is
$f(x)=x^3.1+x^3(-x)+x^3\frac{1}{2!}(-x)^2+x^3\frac{1}{3!}(-x)^3+...+x^3\frac{1}{n!}(-x)^n+...$
$=x^3-x^4+\frac{x^5}{2}x-\frac{x^6}{6}+...+\frac{(-x)^{n+3}}{n!}+...$