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If the polynomial $P(x)=x^3 - 3x^2 -7x -1$ has roots $a,b,c$, find the value of $(\frac{1}{a-b} + \frac{1}{b-c} + \frac{1}{c-a})^2$.

My attempt:

I developed the expression and by Girard I was able to simplify the numerator by finding an integer. But the denominator couldn't

A13235378
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    You can't (unless the value turns out to be zero). The function you want to evaluate is not a symmetric function of the roots. Meaning that Vieta + Newton-Girard is not enough. – Jyrki Lahtonen Jul 05 '22 at 13:06
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    The sum is only invariant under cyclic permutations. If you interchange $a$ and $b$ then its sign changes. Meaning that its square should be symmetric. – Jyrki Lahtonen Jul 05 '22 at 13:12
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    Viewed differently, the discriminant of the polynomial gives you the square $(a-b)^2(b-c)^2(c-a)^2$, and leaves the sign of $(a-b)(b-c)(c-a)$ undetermined. Without any extra information it is impossible to tell the sign, again because it depends on labeling of the roots. – Jyrki Lahtonen Jul 05 '22 at 13:22
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    I had forgotten the square of the expression, now it can be solved? – A13235378 Jul 05 '22 at 14:19
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    Yes, it can. You get the discriminant in the denominator, and the numerator is a symmetric polynomial. Did you work it out? – Jyrki Lahtonen Jul 05 '22 at 15:17
  • Not yet, I still lack knowledge. How do I use discriminant in the denominator? – A13235378 Jul 05 '22 at 18:29

1 Answers1

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by Girard I was able to simplify the numerator by finding an integer

This leaves the denominator to calculate, which, as noted in a comment, is actually the discriminant of the cubic (for verification, its value is $1300$ per WA). The following is a shortcut to calculate the denominator without using the cubic discriminant formula (or WA).

Completing the cube, $\,P(x) = x^3 - 3x^2 \color{red}{+ 3 x - 3 x} -7x -1= (x-1)^3 - 10(x-1) - 10\,$.

Let $\,Q(t) = P(t+1) = t^3 - 10 t - 10\,$, then the roots of $\,Q(t)\,$ are $\,\alpha=a-1\,$,$\,\beta=b-1\,$,$\,\gamma=c-1\,$, so the denominator $\,(a-b)^2(b-c)^2(c-a)^2 = (\alpha-\beta)^2(\beta-\gamma)^2(\gamma-\alpha)^2\,$.

Subtracting $\,\alpha^3 - 10 \alpha - 10 = 0\,$ and $\,\beta^3 - 10 \beta- 10 = 0\,$ gives:

$$ 0 = \alpha^3-\beta^3 - 10(\alpha-\beta) = (\alpha-\beta)(\alpha^2+\alpha\beta+\beta^2 -10) $$

The first factor is non-zero since $\,Q(t)\,$ has no multiple roots, so $\,\alpha \ne \beta\,$, then:

$$ \alpha^2+\alpha\beta+\beta^2-10 = 0 \quad\iff\quad (\alpha-\beta)^2 = 10 - 3\alpha\beta = 10 - \frac{30}{\gamma} $$

Then, using that $\,\alpha+\beta+\gamma=0\,$, $\,\alpha\beta+\alpha\gamma+\beta\gamma=-10\,$, $\,\alpha\beta\gamma = 10\,$ by Vieta's relations:

$$ \require{cancel} \begin{align} (\alpha-\beta)^2(\beta-\gamma)^2(\gamma-\alpha)^2 &= \left(10 - \frac{30}{\alpha}\right)\left(10 - \frac{30}{\beta}\right)\left(10 - \frac{30}{\gamma}\right) \\ &= 1000 - 3000\left(\frac{1}{\alpha}+ \frac{1}{\beta} + \frac{1}{\gamma}\right) \\ &\quad\quad + \cancel{9000\left(\frac{1}{\alpha\beta} + \frac{1}{\beta\gamma} + \frac{1}{\gamma\alpha}\right)} - \frac{27000}{\alpha\beta\gamma} \\ &= 1000 - 3000 \cdot \frac{-10}{10} - \frac{27000}{10} \\ &= 1300 \end{align} $$

dxiv
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