Answer
${x^3} - x = - 6 + 11\left( {x + 2} \right) - 6{\left( {x + 2} \right)^2} + {\left( {x + 2} \right)^3}$
Work Step by Step
We compute the derivatives of $f\left( x \right) = {x^3} - x$ and list them in the table below:
$\begin{array}{*{20}{c}}
n&{{f^{\left( n \right)}}\left( x \right)}&{\dfrac{{{f^{\left( n \right)}}\left( x \right)}}{{n!}}}&{\dfrac{{{f^{\left( n \right)}}\left( { - 2} \right)}}{{n!}}}\\
0&{{x^3} - x}&{{x^3} - x}&{ - 6}\\
1&{3{x^2} - 1}&{3{x^2} - 1}&{11}\\
2&{6x}&{3x}&{ - 6}\\
3&6&1&1\\
4&0&0&0
\end{array}$
By Theorem 1 of Section 11.7, the Taylor series centered at $c=-2$ is given by
$T\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{f^{\left( n \right)}}\left( { - 2} \right)}}{{n!}}{\left( {x + 2} \right)^n}$
Using the table above, we get
$T\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{f^{\left( n \right)}}\left( { - 2} \right)}}{{n!}}{\left( {x + 2} \right)^n} = - 6 + 11\left( {x + 2} \right) - 6{\left( {x + 2} \right)^2} + {\left( {x + 2} \right)^3}$
So, the Taylor series of $f\left( x \right) = {x^3} - x$ centered at $c=-2$ is
${x^3} - x = - 6 + 11\left( {x + 2} \right) - 6{\left( {x + 2} \right)^2} + {\left( {x + 2} \right)^3}$
