The geometrical identity

We call geometrical identity, or Bernouilli's formula, the following expression:

$$a^n - b^n = (a-b) \sum_{p=0}^{n-1} a^{n-p-1}b^p $$

Demonstration

1. By successive divisions

Performing the successive Euclidian divisions of the polynomial \( (a^n - b^n) \) by \( (a - b) \), we notice that:

$$$$	$$ a-b $$
$$\hspace{0.6em} a^n - b^n $$	$$ \textcolor{#446e4f}{a^{n-1}} $$
$$ \textcolor{#8E5B5B}{-a^n} \textcolor{#446e4f}{+a^{n-1}b} - b^n $$	$$ a^{n-1} \textcolor{#446e4f}{+a^{n-2}b} $$
$$ \hspace{1em} 0 \hspace{0.6em} \textcolor{#8E5B5B}{-a^{n-1}b} \textcolor{#446e4f}{+a^{n-2}b^2} - b^n $$	$$ a^{n-1} +a^{n-2}b \textcolor{#446e4f}{+a^{n-3}b^2} $$
$$ \hspace{4.2em} 0 \hspace{1.2em} \textcolor{#8E5B5B}{- a^{n-2}b^2} \textcolor{#446e4f}{+ \dots + a^{n-p}b^{p} + \dots} - b^n $$	$$ a^{n-1} + a^{n-2}b +a^{n-3}b^2 \textcolor{#446e4f}{+ \dots + a^{n-p-1} b^{p}} $$
$$ \hspace{8.2em} 0 \hspace{1.6em} + \dots \textcolor{#8E5B5B}{-a^{n-p}b^{p}} \textcolor{#446e4f}{+ a^{n-p+1}b^{p+1} } + \dots - b^n $$	$$ a^{n-1} + a^{n-2}b +a^{n-3}b^2 + \dots + a^{n-p-1} b^{p} \textcolor{#446e4f}{ + a^{n-p} b^{p+1}} $$
$$ \hspace{15.2em} 0 \hspace{1.4em} \textcolor{#8E5B5B}{- a^{n-p+1}b^{p+1}} + \dots - b^n $$	$$ a^{n-1} + a^{n-2}b +a^{n-3}b^2 + \dots + a^{n-p-1} b^{p} + a^{n-p} b^{p+1} $$
$$ \hspace{20.4em} \dots until \ (p = n - 3)$$	$$ $$
$$ \hspace{20.4em} 0 \hspace{2em} \textcolor{#446e4f}{ + a^2 b^{n-2}} - b^n $$	$$ a^{n-1} + a^{n-2}b +a^{n-3}b^2 + \dots + a^{n-p-1} b^{p} + a^{n-p} b^{p+1} + \dots \textcolor{#446e4f}{+ ab^{n-2} } $$
$$ \hspace{23em} \textcolor{#8E5B5B}{-a^2 b^{n-2}} \textcolor{#446e4f}{+ a b^{n-1}} - b^n $$	$$ a^{n-1} + a^{n-2}b +a^{n-3}b^2 + \dots + a^{n-p-1} b^{p} + a^{n-p} b^{p-1} + \dots + ab^{n-2} \textcolor{#446e4f}{+ b^{n-1} } $$
$$ \hspace{27.4em} \textcolor{#8E5B5B}{- a b^{n-1}} \textcolor{#8E5B5B}{+ b^n } $$	$$ a^{n-1} + a^{n-2}b +a^{n-3}b^2+ \dots + a^{n-p-1} b^{p} + a^{n-p} b^{p-1} + \dots + ab^{n-2} + b^{n-1} $$
$$ \hspace{30em} 0 $$	$$ a^{n-1} + a^{n-2}b +a^{n-3}b^2+ \dots + a^{n-p-1} b^{p} + a^{n-p} b^{p-1} + \dots + ab^{n-2} + b^{n-1} $$

Therefore we finally obtain,

$$a^n - b^n = (a-b) \Bigl[a^{n-1} + a^{n-2}b +a^{n-3}b^2+ \dots + a^{n-p-1} b^{p} + a^{n-p} b^{p-1} + \dots + ab^{n-2} + b^{n-1} \Bigr] $$



And as a result of it,

$$\forall n \in \mathbb{N}, \enspace \forall (a, b) \in \hspace{0.05em} \mathbb{R}^2,$$

$$a^n - b^n = (a-b) \sum_{p=0}^{n-1} a^{n-p-1}b^p $$

Examples