Let $a$ and $b$ be two real numbers. We set $s=a+b$ and $p=ab$. Express $a^{3}+b^{3}$ in terms of $s$ and $p$ only.
Let's calculate:

$$
a^{3}+b^{3}=(a+b)^{3}-3 a^{2} b-3 a b^{2}=(a+b)^{3}-3 a b(a+b)=s^{3}-3 s p .
$$

## Bernoulli's Equality

& Explanation: We notice the following identities:

$$
\begin{aligned}
& a^{1}-b^{1}=(a-b) \times 1 \\
& a^{2}-b^{2}=(a-b) \times(a+b) \\
& a^{3}-b^{3}=(a-b) \times\left(a^{2}+a b+b^{2}\right)
\end{aligned}
$$

We can then intuit the following result:

## Proposition 1.1.4.

Let $a, b$ be real numbers and $n$ a natural number greater than or equal to 1. Then:

$$
a^{n}-b^{n}=(a-b)\left(a^{n-1}+a^{n-2} b+a^{n-3} b^{2}+\cdots+a b^{n-2}+b^{n-1}\right)
$$

Proof. It suffices to expand the factored form, and then a series of simplifications leads to the desired result.