Answer
$$T_{3}(x) =-5(x+2)+3(x+2)^{3}$$
Work Step by Step
Given $$f(x)=3(x+2)^{3}-5(x+2), \quad T_{3}, \quad a=-2$$
Since
\begin{align*}
f(x)&=3(x+2)^{3}-5(x+2)\ \ \ \ \ \ \ &f(-2)=0\\
f'(x)&=9(x+2)^{2}-5\ \ \ \ \ \ \ &f(-2)=-5\\
f''(x)&=18(x+2)\ \ \ \ \ \ \ &f(-2)=0\\
f'''(x)&=18\ \ \ \ \ \ \ &f(-2)=18
\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(-2)+f^{\prime}(-2)(x+2)+\frac{f^{\prime \prime}(-2)}{2 !}(x+2)^{2}+\frac{f^{\prime \prime \prime}(-2)}{3 !}(x+2)^{3} \\
&=-5(x+2) +\frac{18}{3 !}(x+2)^{3} \\
&=-5(x+2)+3(x+2)^{3}
\end{aligned}
