$abc = k$

$ab + bc + ac = l$

$a+b+c = m$

($k$, $l$ and $m$ are known. $a\geq b\geq c$, $x,y,z\in\mathbb{R}$)

Is it possible to solve this equation system for $a$, $b$ and $c$ by only manipulating the equations (squaring, cubing, adding, subtracting, cube rooting...)?

I'm sure that there's a way, here's a quote:

In general, polynomial equations higher than fourth degree are incapable of algebraic solution in terms of a finite number of additions, subtractions, multiplications, divisions, and root extractions. This was also shown by Ruffini in 1813 (Wells 1986, p. 59).

  • 2
    $\begingroup$ To solve which equation system? You mean, $abc=n$,$a+b+c=m$ and $a+b+c=k$ over the integers? $\endgroup$ May 31, 2017 at 12:40
  • $\begingroup$ @MCCCS: Are there any restrictions on $a,b,c$ other than $a,b,c \in \mathbb{R}$, $a \ge b \ge c$, and the given equations? $\endgroup$
    – quasi
    May 31, 2017 at 12:51
  • $\begingroup$ @quasi a, b, c are complex. $\endgroup$
    – MCCCS
    May 31, 2017 at 12:54
  • 3
    $\begingroup$ For complex $a,b,c$ inequalities $a\ge b\ge c$ do not make sense. $\endgroup$ May 31, 2017 at 12:55

2 Answers 2


In general, I don't see how to avoid a cubic for complex solutions. For example, solve the system \begin{align*} abc & = 1,\\ ab+bc+ca & = 2,\\ a+b+c & = 3. \end{align*} The solutions are given by the roots of the cubic $$ c^3-3c^2+2c-1=0, $$ and quadratic equations for $a$ and $b$. Whatever we do otherwise, must involve these roots of the cubic.

  • $\begingroup$ @MCCCS: Said differently: If there was another method for solving such systems of equations, that would yield another method for solving cubic equations. $\endgroup$
    – quasi
    May 31, 2017 at 13:02

Certainly not.

If that was possible, we would have an alternative method of solving a cubic equation.

In a way, solving a cubic equation without solving a cubic equation :)

And as it seems that the values are real (you can order them), you cannot do without the trigonometric approach, this is a "casus irreductibilis".

  • $\begingroup$ As the question was updated by the OP, the last paragraph is now irrelevant. $\endgroup$
    – user65203
    May 31, 2017 at 13:06

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