The general form of the binomial theorem discovered by Newton, that works for any number #p#, can be written:
#(1+y)^{p}=1+py+(p(p-1))/(2!) y^2+(p(p-1)(p-2))/(3!)y^3+(p(p-1)(p-2)(p-3))/(4!)y^4+cdots# for #|y|<1#.
When #p# is a non-negative integer, the infinite series above "truncates" to a finite expansion that is the more familiar binomial theorem from precalculus. Such an expansion is then valid for all #y#.
For the given problem, #y=-2x# and #p=1/3#, so
#(1-2x)^(1/3)=1+(1/3)(-2x)+(1/3 * -2/3)/(2!)(-2x)^2+(1/3 * -2/3 * -5/3)/(3!)(-2x)^3+(1/3 * -2/3 * -5/3 * -8/3)/(4!)(-2x)^4+(1/3 * -2/3 * -5/3 * -8/3 * -11/3)/(2!)(-2x)^5+cdots#
#=1-2/3 x-4/9 x^2-40/81 x^3-160/243 x^4-704/729 x^5-cdots#
This is valid for #|2x|<1 leftrightarrow |x|<1/2#
