Polynomials and Girard 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


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