Using Vieta's theorem for cubic equations to derive the cubic discriminant

Background:

Vieta's Theorem for cubic equations says that if a cubic equation $x^3 + px^2 + qx + r = 0$ has three different roots $x_1, x_2, x_3$, then

$$\begin{eqnarray*} -p &=& x_1 + x_2 + x_3 \\ q &=& x_1x_2 + x_1x_3 + x_2x_3 \\ -r &=& x_1x_2x_3 \end{eqnarray*}$$

The exercise is:

A cubic equation $x^3 + px^2 + qx + r = 0$ has three different roots $x_1, x_2, x_3$. Find $(x_1 - x_2)^2 (x_2 - x_3)^2 (x_1 - x_3)^2$ as an expression containing $p, q, r$.

Spoiler alert: the answer is $-4p^3r - 4q^3 + p^2q - 27r^2 - 18pqr$.

My question is: how am I supposed to find that without using a computer?

• This is called "Vieta's theorem"? I thought this was just "expanding the product of three linear factors". Where'd you get that theorem name? – Patrick Da Silva Jan 29 '12 at 5:09
• This is from a book called "Algebra" by Gelfand/Shen. – Matt Gregory Jan 29 '12 at 5:14
• @PatrickDaSilva: mathworld.wolfram.com/VietasFormulas.html (I'd heard the name before, but I generally think of it as "theory of equations" or "relationship between polynomial coefficients and sums of products of polynomial roots $k$ at a time" or some such) – Isaac Jan 29 '12 at 5:59
• Also see Newton's identities and Vieta's formulas. – Sasha Jan 29 '12 at 6:55

The tedious way is to expand $(x_1 - x_2)^2 (x_2 - x_3)^2 (x_1 - x_3)^2$ out completely and then write it in terms of $x_1 + x_2 + x_3$ and so forth. You are guaranteed that this is possible by the fundamental theorem of symmetric polynomials, the proof of which even gives an algorithm for doing this, but it's a pain to do by hand (although it's not a bad exercise in algebraic manipulation).

A less tedious way is to argue as follows. We will first work under the assumption that $p = 0$. Now, $q, r$ are polynomials of degrees $2, 3$, and the discriminant is a polynomial of degree $6$, so the discriminant must be a linear combination of the two monomials $q^3, r^2$. Thus we can write $$\Delta = a q^3 + b r^2$$

for two constants $a, b$, where $\Delta$ is the discriminant. Setting $q = -1, r = 0$ we get the polynomial $x^3 - x = 0$ with roots $0, 1, -1$. We compute that the discriminant is equal to $4$, from which it follows that $a = -4$.

Setting $q = 0, r = -1$ we get the polynomial $x^3 - 1 = 0$ with roots $1, \omega, \omega^2$ where $\omega$ is a primitive third root of unity. Using the identity $$(\omega - 1)^2 = \omega^2 - 2 \omega + 1 = - 3 \omega$$

we compute that the discriminant is equal to $-27$, from which it follows that $b = -27$. Thus $$\Delta = -4 q^3 - 27 r^2.$$

To get from here to an arbitrary choice of $x_1, x_2, x_3$, apply the above formula to the polynomial with roots $x_1 - \frac{p}{3}, x_2 - \frac{p}{3}, x_3 - \frac{p}{3}$ and note that subtracting the same constant from each of the three roots doesn't change the discriminant.

• My memory is failing me, but I believe I learned this argument from David Speyer somewhere (possibly here). – Qiaochu Yuan Jan 29 '12 at 6:21
• This is really cool. Thank you Qiaochu! – Matt Gregory Jan 29 '12 at 13:07
• Can you please tell me, how did you conclude that q and r are polynomials of degrees 2 and 3? And how did you get x^3-x=0 in the 3rd paragraph? I really don't understand. – Pavel Jul 7 '16 at 19:04