Answer
$$T_3(x) =1+3(x-1)+3(x-1)^{2}+(x-1)^{3}$$
Work Step by Step
Given $$f(x)=x^{3}, \quad T_{3}, \quad a=1$$
Since
\begin{align*}
f(x)&=x^{3}\ \ \ \ \ \ \ &f(1)=1\\
f'(x)&=3x^{2}\ \ \ \ \ \ \ &f(1)=3\\
f''(x)&=6x\ \ \ \ \ \ \ &f(1)=6\\
f'''(x)&=6\ \ \ \ \ \ \ &f(1)=6
\end{align*}
Then
\begin{aligned}
T_{3}(x)&= f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\frac{f^{\prime \prime \prime}(1)}{3 !}(x-a)^{3} \\
&=f(1)+f^{\prime}(1)(x-1)+\frac{f^{\prime \prime}(1)}{2 !}(x-1)^{2}+\frac{f^{\prime \prime \prime}(1)}{3 !}(x-1)^{3} \\
&=1+3(x-1)+\frac{6}{2 !}(x-1)^{2}+\frac{6}{3 !}(x-1)^{3} \\
&=1+3(x-1)+3(x-1)^{2}+(x-1)^{3}
\end{aligned}
