Working through Ian Stewart's "Galois Theory, Third Edition," he states at the end of the second paragraph on page 13: "Because we know that $\alpha_1+\alpha_2+\alpha_3$ is minus the coefficient of $t^2$..." in reference to the monic cubic $P(t)=t^3+a t^2 + b t + c$.

It's easy to verify this just grinding through Cardano's formula solution or with Mathematica:

P[t_] := t^3 + a t^2 + b t + c
r = Solve[P[t] == 0, t];
(r[[1, 1, 2]] + r[[2, 1, 2]] + r[[3, 1, 2]]) // FullSimplify

produces $-a$.

But that's hardly adequate. Stewart states this so offhandedly that he must mean it to be completely obvious, but I don't quite see it. I've been playing around with the Tschirnhaus transformation at the top of page 8 for some time, and I'm convinced that it holds the key to the erstwhile obvious knowledge because it makes the quadratic term go away:

In[..]:= P[y - a/3] // Collect[#, y] &
Out[..]= (2 a^3)/27 - (a b)/3 + c + (-(a^2/3) + b) y + y^3

$P\left(y-\frac{a}{3}\right)=y^3+y \left(b-\frac{a^2}{3}\right)+\frac{2 a^3}{27}-\frac{a b}{3}+c$

Substituting the roots doesn't produce anything more obvious to me.

I'm sure I'm just being obtuse, but I'd be grateful for a ray of light on this.

  • 5
    $\begingroup$ factor the cubic as $(t- \alpha_1)(t - \alpha_2)(t-\alpha_3)$ Now look for terms in the product which are quadratic in t. This works for any polynomial with lead coefficient 1: the coefficient of the second highest degree term is -(sum of the roots) $\endgroup$
    – lulu
    Jul 6, 2015 at 3:26
  • 3
    $\begingroup$ See Vieta's formulas at Wikipedia or elsewhere. $\endgroup$ Jul 6, 2015 at 6:37

1 Answer 1


You can learn this by expanding the factored form of the cubic. If we suppose $a$, $b$ and $c$ are the roots, then: $$(x-a)(x-b)(x-c)=x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc$$ It's a good thought to do this out by hand if you've not seen it before. Notice that when we multiply monomials like $(x-a)$, we just "take" one term from each monomial and multiply them together - so the coefficients of $x^2$ are when we "take" the $x$ from $2$ of the monomials and take the (negative) constant from the remaining one. Summing all that up yields $-(a+b+c)$ as the coefficient of $x^2$.


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