#### Answer

\begin{aligned}
T_{2}(x) &=75+106(x-3)+54(x-3)^{2} \\
T_{3}(x) &=75+106(x-3)+54(x-3)^{2}+12(x-3)^{3}
\end{aligned}

#### Work Step by Step

Since
\begin{array}{ll}
{f(x)=x^{4}-2 x,} & {f(3)=75} \\
{f^{\prime}(x)=4 x^{3}-2,} & {f^{\prime}(3)=106} \\
{f^{\prime \prime}(x)=12 x^{2},} & {f^{\prime \prime}(3)=108} \\
{f^{\prime \prime \prime}(x)=24 x,} & {f^{\prime \prime \prime}(3)=72}
\end{array}
Then
\begin{aligned}
T_{2}(x) &=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2} \\
&=75+106(x-3)+54(x-3)^{2} \\
T_{3}(x) &=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\frac{f^{\prime \prime \prime}(a)}{3 !}(x-a)^{3} \\
&=75+106(x-3)+54(x-3)^{2}+12(x-3)^{3}
\end{aligned}