Answer
$$T_{3}(x) = 2+\frac{1}{4}(x-2)-\frac{1}{32}(x-2)^{2}+\frac{5}{768}(x-2)^{3}$$
Work Step by Step
Given $$f(x)=(3 x+2)^{1 / 3}, \quad T_{3}, \quad a=2$$
Since
\begin{align*}
f(x)&=(3 x+2)^{1 / 3},\ \ \ \ \ \ \ &f(2)=2\\
f'(x)&=(3 x+2)^{-2 / 3},\ \ \ \ \ \ \ &f(2)=1/4\\
f^{\prime \prime}(x) &=-2(3 x+2)^{-\frac{3}{3}} \quad &f^{\prime \prime}(2)=-\frac{1}{16}\\
f^{\prime \prime \prime}(x) &=10(3 x+2)^{-\frac{8}{3}} \quad &f^{\prime \prime \prime}(2)=\frac{5}{128}
\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}(a)}{3 !}(x-a)^{3} \\
&= 2+\frac{1}{4}(x-2)-\frac{1}{32}(x-2)^{2}+\frac{5}{768}(x-2)^{3}
\end{aligned}
