Let $P=X^{3}-3 X-1$. Calculate the sum of the inverses of the squares of the roots (real or complex) of $P$.

Soit $P=X^{3}-3 X-1$. Calculer la somme des inverses des carrés des racines (réelles ou complexes) de $P$.

(Note: The second line is kept in French as per the original text, but the first line is translated to English.)
Let's denote $r_{1}, r_{2}, r_{3}$ as these roots. Then by Viète's formulas:

$$
\begin{aligned}
\frac{1}{r_{1}^{2}}+\frac{1}{r_{2}^{2}}+\frac{1}{r_{3}^{2}} & =\left(\frac{1}{r_{1}}+\frac{1}{r_{2}}+\frac{1}{r_{3}}\right)^{2}-2 \cdot\left(\frac{1}{r_{2} r_{3}} \frac{1}{r_{1} r_{2}} \frac{1}{r_{1} r_{3}}\right) \\
& =\left(\frac{r_{1} r_{2}+r_{1} r_{3}+r_{2} r_{3}}{r_{1} r_{2} r_{3}}\right)^{2}-2 \cdot\left(\frac{r_{1}+r_{2}+r_{3}}{r_{1} r_{2} r_{3}}\right) \\
& =(-3)^{2}-2 \cdot 0=9
\end{aligned}
$$

However, it is not possible to express the roots simply in terms of the coefficients in the general case (only for degrees $1,2,3,4$, and it is impossible for degrees greater than 5 (Galois groups)).

The Lagrange interpolation polynomial answers the question: can we find a polynomial whose graph passes through certain points?

Take $n$ distinct real numbers $x_{1}, \ldots, x_{n}$ and $n$ real numbers $y_{1}, \ldots, y_{n}$.

The initial idea is to start by solving an easier problem: find a polynomial $L_{k}$ of degree $n-1$ that vanishes at $n-1$ distinct real numbers $x_{1}, \ldots, x_{k-1}, x_{k+1}, \ldots, x_{n}$ and takes a certain value $y_{k}$ at $x_{k}: L_{k}=y_{k} \cdot \prod_{i \neq k} \frac{X-x_{i}}{x_{k}-x_{i}}$ works. It suffices to sum these polynomials to obtain the desired result.

The interpolation polynomial of degree $n-1$ is unique. It follows that the polynomial solutions are exactly the

$$
Q(X) \cdot \prod_{k=1}^{n}\left(X-x_{k}\right)+\sum_{k=1}^{n} L_{k}(X)
$$

with $Q(X)$ any polynomial.

All these results also hold for rationals and complexes.