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The problem was asked before: Using Vieta's theorem for cubic equations to derive the cubic discriminant . I tried to solve it by purely algebraic manipulation but was faced with an explosion of terms too confusing to simplify. I have given up trying and accepted the fact at face value but would like to know if vieta himself did this by mechanical manipulation or some other method? Can someone show me the actual derivation completely?(I could not find it on google)

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Aditya Chintalapati
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2 Answers2

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I happen to have an explicit calculation written down, so here goes.

The key result to be familiar with is the fundamental theorem of symmetric polynomials. By this I mean not the result but with the process of converting any symmetric polynomial into expressions in the elementary symmetric polynomials. In our case, the coefficients of the given cubic are $-\sigma_1$, $\sigma_2$, and $-\sigma_3$, via the expansion of $(X-x_1)(X-x_2)(X-x_3)$.

I'm afraid that this will get (even more) unwieldy unless I condense notation. I will use $\Sigma x_1^ax_2^bx_3^c$ to refer to the sum of the elements in the orbit of the monomial $x_1^ax_2^bx_3^c$ under the action of the symmetric group. For instance $\Sigma x_1 = x_1 + x_2 + x_3 = \sigma_1$ and $\Sigma x_1^3x_2^2x_3 = x_1^3x_2^2x_3+x_1^3x_2x_3^2+x_1^2x_2^3x_3+x_1^2x_2x_3^3+x_1x_2^3x_3^2+x_1x_2^2x_3^3$.

First, I expand $\Delta$, which is not symmetric:

  • $\Delta = (x_1-x_2)(x_1-x_3)(x_2-x_3) = x_1^2x_2 - x_1^2x_3 - x_2^2x_1 + x_2^2x_3 - x_3^3x_2 + x_3^2x_1$.

    Now we square it! Using the condensed notation I check directly

  • $\Delta^2 = \Sigma x_1^4x_2^2 - 2\Sigma x_1^4x_2x_3 - 2 \Sigma x_1^3x_2^3 +2\Sigma x_1^3x_2^2x_3 - 6 x_1^2x_2^2x_3^2$.

Now, it only remains to write this symmetric polynomial as a polynomial in $\sigma_1, \sigma_2, \sigma_3$. This is pretty tedious, but entirely mechanical. The leading terms justify why I'm choosing these particular expansions. As an aside, I compute only the coefficients of one monoial per orbit, and I also plug in $x_1=x_2=x_3=1$ as a double-check.

  • $\sigma_1^2\sigma_2^2 = \Sigma x_1^4x_2^2 + 2\Sigma x_1^4x_2x_3 + 2\Sigma x_1^3x_2^3 +8\Sigma x_1^3x_2^2x_3+ 15x_1^2x_2^2x_3^2$

  • $\sigma_1^3\sigma_3 = \Sigma x_1^4x_2x_3 + 3\Sigma x_1^3x_2^2x_3 + 6 x_1^2x_2^2x_3^2$

  • $\sigma_2^3 = \Sigma x_1^3x_2^3 + 3\Sigma x_1^3x_2^2x_3 + 6x_1^2x_2^2x_3^2$

  • $\sigma_1\sigma_2\sigma_3 = \Sigma x_1^3x_2^2x_3 + 3x_1^2x_2^2x_3^2$

  • $\sigma_3^2 = x_1^2x_2^2x_3^2$

Now, it's just a matter of finding the right linear combination

  • $\Delta^2 = \sigma_1^2\sigma_2^2 - 4\sigma_1^3\sigma_3 - 4\sigma_2^3 + 18\sigma_1\sigma_2\sigma_3 - 27 \sigma_3^2$.

Replace $\sigma_1,\sigma_2, \sigma_3$, with $-b, c, -d$, and you have the discriminant of $x^3+bx^2+cx+d$.

Rolf Hoyer
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There is a more intuitive way to get the discriminant of a cubic polynomial. A cubic polynomial $p(x)$ will have $3$ roots iff its critical points have opposite sign.

So let $Q_1$, $Q_2$ be the roots of $p'(x)$. The discriminant is the expression $p(Q_1)p(Q_2)$ whose denominator has been cleared.

trying
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Jimmy Dillies
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  • I like this answer (it's nice to have an easy way to recall the expression). But the correct normalization is $\mathrm{Disc}(p) = -27p(Q_1)p(Q_2)$, so it's not enough to clear denominators (how would you get the sign?). Maybe one can use an easy polynomial to compute the normalization with, like $p(x)=x^3 - 3x$ (the roots are $0$ and $\pm \sqrt{3}$, and the roots of $p'$ are $\pm 1$. So one can easily compute both the discriminant in the form $\prod_{i<j}(r_i - r_j)^2$ and the expression above. Here the $r_i$ are the roots of $p$.) – Geva Yashfe Jun 19 '25 at 22:09