<s>[INST] 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.
 [/INST] We have for $x\neq 0$
$$\dfrac{1}{x} = 1 + (1 - x) + (1 - x)^2 + (1 - x)^3/x$$
$$ (1 - x)^3 + x(1 - x)^2\color{red}{+ x(1 - x)-(1-x)}=0 $$
$$(1 - x)^3 + x(1 - x)^2\color{red}{+ (1 - x)(x-1)}=0$$
$$(1 - x)^3 \color{blue}{+ x(1 - x)^2- (1 - x)^2=0}$$
$$(1 - x)^3 \color{blue}{+ (1 - x)^2(x-1)}=0$$
$$(1 - x)^3 \color{blue}{- (1 - x)^3}=0$$
</s>