. Let $u$ and $v$ be two real numbers. We set $s=u+v, p=u v$. Calculate $u^{3}+v^{3}$ in terms of $s$ and $p$.
. On a

$$
s^{3}=(u+v)^{3}=u^{3}+3 u^{2} v+3 u v^{2}+v^{3}=u^{3}+v^{3}+3 u v(u+v)=u^{3}+v^{3}+3 s p
$$

so $u^{3}+v^{3}=s^{3}-3 s p$.

### 1.3 Cardano's Formula

This method for solving cubic equations, discovered in the 16th century, played an important role in the history of mathematics. We will limit ourselves to a numerical example.

We seek the value of $x$ for which $x^{3}+3 x+6=0$. For this, we use the previous exercise by looking for two numbers $u$ and $v$ such that $u v=-1$ and such that $x=u+v$ is a solution to this equation. We have $x^{3}=(u+v)^{3}=u^{3}+v^{3}+3 u v(u+v)=u^{3}+v^{3}-3 x$ because by hypothesis $u v=-1$. Therefore, $x^{3}+3 x+6=u^{3}+v^{3}+6=u^{3}+\left(\frac{-1}{u}\right)^{3}+6$, or by multiplying by $u^{3}$:

$$
\left(u^{3}\right)^{2}+6 u^{3}-1=0
$$

So $u^{3}$ is a solution to a quadratic equation, and we know how to solve quadratic equations:

$$
\left(u^{3}+3\right)^{2}-10=0
$$

thus $u^{3}=-3+\sqrt{10}$ or $u^{3}=-3-\sqrt{10}$.

But for any real number $y$, there exists a real number $z$ such that $z^{3}=y$, and we denote it as: $z=\sqrt[3]{y}$. Furthermore, since $(-3+\sqrt{10})(-3-\sqrt{10})=-1$, if $u^{3}=-3+\sqrt{10}$, then $v^{3}=-3-\sqrt{10}$ and vice versa (because by hypothesis $u v=-1$). In the end, this method gives a solution to the equation

$$
x=u+v=\sqrt[3]{-3+\sqrt{10}}+\sqrt[3]{-3-\sqrt{10}}
$$

and it can be numerically verified that this number is indeed the only real solution to the equation. The problem this posed is that for certain cubic equations, especially those that admit three roots, the intermediate quadratic equation does not have real roots, and the idea came to Cardano that it still had roots, but "imaginary," and that by passing through these imaginary roots one could find the real roots of the original equation. This is how complex numbers were created, but it took more than two centuries to define them satisfactorily.