Find $(a, b, c)$ real numbers satisfying:

$a+b+c=3, a^{2}+b^{2}+c^{2}=35, a^{3}+b^{3}+c^{3}=99$.

## - Solution -
It is necessary to reduce to a third-degree equation using the fact that any symmetric expression of the roots of an equation can be expressed rationally in terms of the elementary symmetric polynomials (coefficients of the said third-degree equation). In this case, $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(ab+bc+ca)$, so $ab+bc+ca=-13$. And $a^{3}+b^{3}+c^{3}=(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)-\left(a^{2}b+b^{2}a+a^{2}c+c^{2}a+b^{2}c+c^{2}b\right)$ and $\left(a^{2}b+b^{2}a+a^{2}c+c^{2}a+b^{2}c+c^{2}b\right)=(a+b+c)(ab+ac+bc)-3abc$. Hence: $99-3 \times 35=$ $-\left((3 \times -13)-3abc\right)$, hence: $abc=-15$. $a, b$ and $c$ are therefore the roots of: $x^{3}-3x^{2}-13x+15$. This equation has 1 as an obvious root, it factors into: $(x-1)\left(x^{2}-2x-15\right)$ and therefore has the other roots: -3 and 5. Thus $(a, b, c)$ is, up to permutation: $(1, -3, 5)$.

## 2 Second TD on polynomials

## - Statements -