Well, if you don't know any shortcut (neither do I), it might as well be good to practice writing the full expansion out.
The Taylor series expansions are written as:
#sum_(n=0)^(oo) f^n(a)/(n!)(x-a)^n#
where #f^n(a)# is the nth derivative of #f(x)# at #x->a# (except for the #x - a# term, in which #x# remains as #x#). In this case, it doesn't matter what #a# is, because we only want the coefficient. Let's just say #a = 2#.
#f(x) = x^6#
#f'(x) = 6x^5#
#f''(x) = 30x^4#
#f'''(x) = 120x^3#
#f''''(x) = 360x^2#
#f'''''(x) = 720x#
#f''''''(x) = 720#
And past that is just #0#.
#=> f(2)/(0!)(x-2)^(0) + (f'(2))/(1!)(x-2)^(1) + (f''(2))/(2!)(x-2)^(2) + (f'''(2))/(3!)(x-2)^(3) + (f''''(2))/(4!)(x-2)^(4) + (f'''''(2))/(5!)(x-2)^(5) + (f''''''(2))/(6!)(x-2)^(6)#
#= 2^6/(1) + (6*2^5)/1(x-2) + (30*2^4)/(2)(x-2)^2 + (120*2^3)/(6)(x-2)^3 + (360*2^2)/(24)(x-2)^4 + (720*2)/(120)(x-2)^5 + (720)/(720)(x-2)^6#
Coefficients then are:
#color(blue)(64, 192, 240, 160, 60, 12, 1)#
The full expansion is:
#color(blue)(64+ 192(x-2) + 240(x-2)^2 + 160(x-2)^3 + 60(x-2)^4 + 12(x-2)^5 + (x-2)^6)#
