<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] 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:
$$a^3-b^3=(a-b)(a^2+ab+b^2).$$
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