Consider the polynomial:
$$x^3-2x^2+3x+5=0$$
where $x_1, x_2$ and $x_3$ are the roots of the above polynomial. Now, consider the following determinant, which is defined using the above given roots:
$$\Delta = \begin{vmatrix}
1 & 1 & 1 \\
x_1 & x_2 & x_3 \\
x_1^2 & x_2^2 & x_3^2 \\
\end{vmatrix}$$
And what is asked of me is to find $\Delta^2$.
After a bit of manipulation I found the following:
$$\Delta = (x_2-x_1)(x_3-x_1)(x_3-x_2)$$
Interestingly enough, this type of matrix has a special name: Vandermonde matrix and instead of doing that bit of manipulation after which I arrived at the above expression for $\Delta$, I could've used the formula given on that wikipedia page. Anyways…
So, I have to find:
$$\Delta^2 = (x_2-x_1)^2(x_3-x_1)^2(x_3-x_2)^2$$
The problem is that I cannot find any of the roots. I used the rational root theorem and found that there are no rational roots. None of the divisors of the free term, $5$, give $0$ when plugged into the polynomial. I tried all options: $\{\pm 1, \pm 5 \}$ and they all give something $\ne 0$.
So then I used the notation:
$$f(x) = x^3-2x^2+3x+5$$
found the derivative:
$$f'(x) = 3x^2-4x+3$$
and I observed that $f'(x) > 0$ for all $x \in \mathbb{R}$. So the function $f$ is strictly increasing, so we can have at most one solution to $f(x) = 0$. Because of what I showed above, this solution cannot be rational. So I concluded that we have one rational root and two complex (and conjugate, since $f \in \mathbb{R}[X]$) roots. But this is as far as I got. I cannot find them. And I tried finding $\Delta^2$ without finding the roots, but I couldn't solve that either.
Best Answer
Let $x_1+x_2+x_3=3u$, $x_1x_2+a_1x_3+x_2x_3=3v^2$ and $x_1x_2x_3=w^3$.
Thus, $$u=\frac{2}{3},$$ $$v^2=1,$$ $$w^3=-5$$ and $$(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2=$$ $$=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)=$$ $$=27\left(\frac{4}{3}-4+\frac{160}{27}-20-25\right)=-1127.$$