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Use Viete's relations to prove that the roots $x_1$, $x_2$, and $x_3$ of the equation $x^3+ax+b=0$ satisfy the identity $(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2=-4a^3-27b^2$.

I know that viete's relations state that the roots $x_1$, $x_2$, and $x_3$ of the equation $x^3-px^2+qx-r=0$ have the property $p=x_1+x_2+x_3$, $q=x_1x_2+x_1x_3+x_2x_3$ and $r=x_1x_2x_3$.

My question is whether or not there is a way to do this without multiplying out $(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2$ and showing that it factors into $4(x_1+x_2+x_3)^3+27(x_1x_2x_3)$ because that algebra involved in that looks like it will be nasty.

mathlove
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chris
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    This is a simpler duplicate; see https://math.stackexchange.com/questions/103491/using-vietas-theorem-for-cubic-equations-to-derive-the-cubic-discriminant?rq=1 – Hw Chu Dec 13 '17 at 01:09
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    @HwChu That solution uses calculating the discriminant of a cubic function as part of the solution, but I'm supposed to be proving that the discriminant has that formula. – chris Dec 13 '17 at 01:34

3 Answers3

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Let $$f(x)=x^3+ax+b=(x-x_1)(x-x_2)(x-x_3)$$ so that $$f'(x)=3x^2+a=(x-x_1)(x-x_2)+(x-x_1)(x-x_3)+(x-x_2)(x-x_3)$$ This allows us to get the equation $$ f'(x_1)f'(x_2)f'(x_3)=(3x_1^2+a)(3x_2^2+a)(3x_3^2+a)=-(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2 $$ We try to evaluate $$ \begin{align} (x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2 &= -(3x_1^2+a)(3x_2^2+a)(3x_3+a) \\ &=-(a^3+3a^2(x_1^2+x_2^2+x_3^2)+9(x_1^2 x_2^2+x_1^2 x_3^2+x_2^2x_3^2)a+27x_1^2x_2^2x_3^2) \end{align} $$ Using the Vieta's formulas

$$ \begin{align} x_1+x_2+x_3 &= 0\\ x_1x_2+x_1x_3+x_2x_3 &= a\\ x_1x_2x_3 &=-b \end{align} $$ we obtain $$ \begin{align} x_1^2+x_2^2+x_3^2 &= (x_1+x_2+x_3)^2 - 2(x_1x_2+x_1x_3+x_2x_3)\\ &=-2a\\ x_1^2x_2^2+x_1^2x_3^2+x_2^2x_3^2 &= (x_1x_2+x_1x_3+x_2x_3)^2-2x_1x_2x_3(x_1+x_2+x_3)\\ &= a^2 \end{align} $$

Hence we get the final expression $$ (x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2 = -(a^3-6a^3+9a^3+27b^2)=-(4a^3+27b^2) $$

Yong Hao Ng
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If $b=0$, then the roots are $0,\pm\sqrt{-a}$ from which $$(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2=-4a^3$$ follows.

In the following, $b\not=0$.

We have, by Vieta's formulas, $$x_1+x_2+x_3=0,\quad x_1x_2+x_2x_3+x_3x_1=a,\quad x_1x_2x_3=-b$$

So, we can have $$\begin{align}x_3(x_1-x_2)^2&=x_3((x_1+x_2)^2-4x_1x_2)\\\\&=x_3(-x_3)^2-4x_1x_2x_3\\\\&=x_3^3-4(-b)\\\\&=(-ax_3-b)-4(-b)\\\\&=-ax_3+3b\end{align}$$ Similarly, we get $$x_1(x_2-x_3)^2=-ax_1+3b,\qquad x_2(x_3-x_1)^2=-ax_2+3b$$

It follows from these that $$\begin{align}&x_1x_2x_3(x_1-x_2)^2(x_2-x_3)^2(x_3-x_1)^2\\\\&=x_3(x_1-x_2)^2x_1(x_2-x_3)^2x_2(x_3-x_1)^2\\\\&=(-ax_3+3b)(-ax_1+3b)(-ax_2+3b)\\\\&=-a^3x_1x_2x_3+3a^2b(x_1x_2+x_2x_3+x_3x_1)-9ab^2(x_1+x_2+x_3)+27b^3\\\\&=-a^3(-b)+3a^2b\cdot a-9ab^2\cdot 0+27b^3\end{align}$$

Dividing the both sides by $x_1x_2x_3=-b\not=0$ gives $$(x_1-x_2)^2(x_2-x_3)^2(x_3-x_1)^2=-4a^3-27b^2$$

mathlove
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  • Ok, that all makes sense, except I don't see why $b=0$ is easy to deal with. Pretty much the whole process here depends on $b \neq 0$ and I can't find any way to use $b=-x_1x_2x_3=0$ to make the messy algebra less awful. – chris Dec 13 '17 at 22:34
  • @chris: I've added some explanations. – mathlove Dec 14 '17 at 04:56
  • Yeah, that was incredibly simple. I'm kinda angry with myself for not figuring that out. Thanks. – chris Dec 14 '17 at 16:37
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Let's first prove Vieta's cubic relations.

$\underline {\text{Proof}}$: Let the roots of a cubic polynomial, $f(x)$, be $\alpha$, $\beta$, $\gamma$. Then $f(x) = (x-\alpha)(x-\beta)(x-\gamma)$. Let $f(x) = x^3 - px^2 + qx - r$.

\begin{align} f(x) & = (x-\alpha)(x-\beta)(x-\gamma) \\ & = (x^2 - \beta x - \alpha x + \alpha \beta) (x - \gamma) \\ & = (x^3 - \gamma x^2 - \beta x^2 + \beta \gamma x - \alpha x^2 + \alpha \gamma x + \alpha \beta x - \alpha \beta \gamma) \\ & = x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha \beta + \alpha \gamma + \beta \gamma)x - (\alpha \beta \gamma) \\ \implies p & = \alpha + \beta + \gamma \\ q & = \alpha \beta + \alpha \gamma + \beta \gamma \\ r & = \alpha \beta \gamma \end{align}

The proof of a cubic discriminant is quite long and there is no easy way to do it, so I'll link a proof here.

Now with the question.

$\underline {\text{Proof}}$: A cubic polynomial $f(x) = (x-\alpha)(x-\beta)(x-\gamma) = x^3 - px^2 + qx - r$ has a discriminant

\begin{align} \Delta_3 & = (\alpha-\beta)^2 (\alpha-\gamma)^2 (\beta-\gamma)^2 \\ \end{align}

Assume $p=0$. We have that $q$ and $r$ are polynomials of degrees $2$ and $3$. The discriminant is a polynomial of degree $6$ (look at the definition just above), and hence the discriminant must be a linear combination of the one-term polynomials, $q^3$ and $r^2$.

$$\Delta_3 = mq^3 + nr^2$$

where $m$ and $n$ are two constants. Now, we can do something clever to get our final result. Let $q = -1$ and $r=0$. Now we have a new polynomial

\begin{align} 0 & = x^3 - x \\ & = x(x^2 - 1)\\ & = x(x-1)(x+1)\\ \end{align}

with roots $-1, 0, 1$.

\begin{align} \Delta_3 & = (\alpha-\beta)^2 (\alpha-\gamma)^2 (\beta-\gamma)^2 \\ & = (-1 - 0)^2 (-1 - 1)^2 (0 - 1)^2\\ & = 1 \cdot 4 \cdot 1\\ & = 4 \\ \end{align}

So far, if you've been keeping up, we have $\Delta = 4q^3 + br^2$.

In a similar method, if we set $q=0$, $r=−1$ we get the polynomial $x^3−1=0$. Now we are in the territory of complex numbers! Solving for roots of $x^3-1=0$ we get $1, \omega, \omega^2$. $\omega$ is a third root of unity.

\begin{align} \Delta_3 & = (\alpha-\beta)^2 (\alpha-\gamma)^2 (\beta-\gamma)^2 \\ & = (1 - \omega)^2 (1 - \omega^2)^2 (\omega - \omega^2)^2 \\ \end{align}

Finishing this we get the discriminant is equal to $27$.

Hence, we have

\begin{align} \Delta_3 &= 4q^3 + 27r^2 \\ & = 4(\alpha \beta + \alpha \gamma + \beta \gamma)^3 + 27(\alpha \beta \gamma)^2 \\ \end{align}

And we are done.

Landuros
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