Answer
$$ T_{3}(x) =x-2 x^{2}+2 x^{3}$$
Work Step by Step
Given $$f(x) =xe^{-2x},\ \ a=0$$
Since
\begin{align*}
f(x) &=xe^{-2x}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f(0) =0\\
f'(x) &=(1-2x)e^{-2x}\ \ \ \ \ \ \ f'(0 )=1 \\
f''(x) &=4(x-1)e^{-2x}\ \ \ \ \ \ \ f''(0 )=-4\\
f'''(x) &=4(3-2x)e^{-2x}\ \ \ \ \ \ f'''(0 )=12
\end{align*}
Then
\begin{align*}
T_{3}(x)&=\sum_{n=0}^{3} \frac{f^{(n)}(0)}{n !} x^{n}\\
&=\frac{0}{1} \cdot 1+\frac{1}{1} x^{1}+\frac{-4}{2} x^{2}+\frac{12}{6} x^{3}\\
&=x-2 x^{2}+2 x^{3}
\end{align*}
