As nobody answers, I will do it by myself :)
In general it is useful to calculate the Groëbner basis of the system given a certain monomial ordering. Once we have the Groëbner basis G, the following conditions are equivalent:
- For each unknown of the polynomial $x_i$, there is some $m_i ≥ 0$
such that $x_i^{m_i}$ is the leading monomial for some gi ∈ G.
- The set of the solutions V is finite.
Furthermore, we can also bound the number of solutions by the following result:
$$
|V|\leq \prod m_i
$$
where the $m_i$ are the degrees of the leading monomials appearing in G for each unknown $x_i$.
A less strict bound is represented by the Bezout bound, that says that the number of solutions, if finite, is minor or equal to the product of the degrees of the equations of the original system.
This is what I have found; no idea if we can improve these results in some way or if we can find a tight bound without computing the Groëbner basis.
Any further advice is welcome!
PS
All the infos have been taken by the book
"Ideals, Varieties, and Algorithms
An Introduction to Computational Algebraic Geometry and Commutative Algebra"
Authors: Cox, David A, Little, John, O'Shea, Dona
You can use Vieta's formulas. In particular, with an initial coefficient of $1$, i.e.,
$$P(s) = s^3 + a_2 s^2 + a_1 s + a_0 \tag{1}\label{eq1}$$
has $a_0 = -xyz = -18$, $a_1 = xy+yz+xz = -6$ and $a_2 = -(x + y + z) = 3$. Thus, you have
$$P(s) = s^2 + 3s^2 - 6s - 18 \tag{2}\label{eq2}$$
You could use this to solve for the individual roots using the Cubic function formula.
Update: As discussed in the question comments and the comments below, it's often easier & faster, especially for simpler type cubic equations (e.g., all coefficients are integers) to first try other methods to determine one root, e.g., by factoring by grouping (see How to Factor by Grouping for details) or using the Rational root theorem. If you do determine a root, you can then reduce your sets of equations to a quadratic, so you can then use the Quadratic formula to get the other $2$ roots. If you can't find an initial root, note the suggestion above of using the cubic function formula will always work.
Best Answer
The fundamental theorem of algebra states that any polynomial will have the same number of roots as the degree of the polynomial (counting repeated roots).
Looking at the last set of equations, we can use Viète's formulas to find the cubic which solves the original set of equations.
$x + y + z = \frac{-b}{a}$, $xy + xz + yz = \frac{c}{a}$, $xyz = \frac{-d}{a}$, therefore plugging in the values from the last set of equations gives you: $3a = -b, 3a = c, a = -d$.
Knowing that the general form of a cubic is $ax^3 + bx^2 + cx + d$ and using the relations, the cubic used to solve the last set of equations is $ax^3-3ax^2+3a-a=0$. Since $a \neq 0$, $x^3 - 3x^2 + 3x - 1 = 0$.
Therefore, the cubic is guaranteed to have $3$ roots (including repeated roots) because of the fundamental theorem of algebra and hence, the last set of equations.