Given the first three elementary symmetric polynomials $e_1 = x_1 + x_2 + x_3$, $e_2 = x_1 x_2 + x_1 x_3 + x_2 x_3$, $e_3 = x_1 x_2 x_3$ (where $x_1, x_2, x_3 \ge 0$), I wish to solve these equations to express $x_1, x_2, x_3$ in terms of $e_1, e_2, e_3$ as simply as possible.
As long as $e_1, e_2, e_3 \ge 0$, a unique (non-negative) solution can be shown to exist by eliminating any two variables, say $x_2, x_3$, to obtain a cubic equation for the remaining variable of the form $x_1^3 - e_1 x_1^2 + e_2 x_1 - e_3$, which by Descarte's rule of signs should have 1 or 3 positive solutions, and also (via the substitution $x_1 \rightarrow -x_1$) no negative solutions, so this cubic always has a single positive solution. Thus the equations themselves should always have a unique non-negative solution, which could be obtained from Cardano's formula or via the usual trigonometric formula.
The question is "is there a simple and symmetric way of expressing the unique solution to this system of equations that does not explicitly use complex numbers or trigonometric functions?"