Proving an equation holds for $x \neq 0$ I would like to show that for $x \neq 0$,
$\dfrac{1}{x} = 1 + (1 - x) + (1 - x)^2 + (1 - x)^3/x.$
One way would be to just expand everything, but is there an easier way?
The sum of a geometric series is $1/(1 - y)$ Letting $y = 1 - x,$ we see
$1/x = \sum_{n = 0}^{\infty} (1 - x)^{n}$
So this takes care of the first terms, but then I just need to show 
$\sum_{n = 3}^{\infty} (1 - x)^{n} = (1 - x)^{3}/x.$
Or, maybe I'm approaching this completely incorrectly.

Notice that the infinite geometric sum is given as following $$a+aq+aq^2+aq^3+\cdots=\dfrac{a}{1-q}$$therefore $$1+(1-x)+(1-x)^2+(1-x)^3+\cdots=\dfrac{1}{1-(1-x)}={1\over x}$$and $$(1-x)^3+(1-x)^4+(1-x)^5+\cdots=\dfrac{(1-x)^3}{x}$$so we have $$1+(1-x)+(1-x)^2+(1-x)^3+\cdots=1+(1-x)+(1-x)^2+(1-x)^3+(1-x)^4+(1-x)^5+\cdots=1+(1-x)+(1-x)^2+\dfrac{(1-x)^3}{x}=\dfrac{1}{x}$$for $0<x<2$
Alternative way: use the following equality $$a^3-b^3=(a^2+ab+b^2)(a-b)$$with $a=1$ and $b=1-x$
