The general form of the binomial theorem discovered by Newton, that works for any number pp, 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(1+y)p=1+py+p(p−1)2!y2+p(p−1)(p−2)3!y3+p(p−1)(p−2)(p−3)4!y4+⋯ for |y|<1|y|<1.
When pp 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 yy.
For the given problem, y=-2xy=−2x and p=1/3p=13, 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
