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.
 A: 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 $b=1-x$
A: Alternatively:
$$\dfrac{1}{x} = 1 + (1 - x) + (1 - x)^2 + \frac{(1 - x)^3}{x} \iff \\
\dfrac{1}{x} = \frac{(1-(1-x))(1 + (1 - x) + (1 - x)^2)}{1-(1-x)} + \frac{(1 - x)^3}{x} \iff \\
\dfrac{1}{x} = \frac{1^3-(1-x)^3}{x} + \frac{(1 - x)^3}{x},$$
where it was used:
$$a^3-b^3=(a-b)(a^2+ab+b^2).$$
A: We have for $x\neq 0$
$$\dfrac{1}{x} = 1 + (1 - x) + (1 - x)^2 + (1 - x)^3/x$$
$$ (1 - x)^3 + x(1 - x)^2\color{red}{+ x(1 - x)-(1-x)}=0 $$
$$(1 - x)^3 + x(1 - x)^2\color{red}{+ (1 - x)(x-1)}=0$$
$$(1 - x)^3 \color{blue}{+ x(1 - x)^2- (1 - x)^2=0}$$
$$(1 - x)^3 \color{blue}{+ (1 - x)^2(x-1)}=0$$
$$(1 - x)^3 \color{blue}{- (1 - x)^3}=0$$
