There are two methods to get the answer.

1) Use the expression #f(-2)+f'(-2)(x+2)+(f''(-2))/(2!)(x+2)^2+(f'''(-2))/(3!)(x+2)^3+\cdots#

Since #f(x)=x-x^3#, we get #f(-2)=-2+8=6#, #f'(x)=1-3x^2# so that #f'(-2)=1-12=-11#, #f''(x)=-6x# so that #f''(-2)=12#, and #f'''(x)=-6# so that #f'''(-2)=-6#. All higher-order derivatives are zero since #f# is a cubic.

The answer is therefore #6-11(x+2)+6(x+2)^2-(x+2)^3#.

2) Let #u=x+2# so that #x=u-2# and expand #f(x)=f(u-2)# before replacing #u# with #x+2# at the end.

Here are the details, which follow from the binomial theorem (Pascal's triangle ).

#f(u-2)=(u-2)-(u-2)^3#

#=u-2-(u^3-6u^2+12u-8)=6-11u+6u^2-u^3#.

The answer is therefore, as we saw before,

#6-11(x+2)+6(x+2)^2-(x+2)^3#.