Given that $x_1, x_2, x_3$ are the roots of the polynomial $x^3-2x^2+3x+5=0$ find $(x_2-x_1)^2(x_3-x_1)^2(x_3-x_2)^2$. 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.
 A: $\Delta$ is not symmetric, but $\Delta^2$ is, so it can be expressed in terms of $a=x_1+x_2+x_3$, $b=x_1x_2+x_2x_3+x_3x_1$ and $c=x_1x_2x_3$. Indeed, we have:
$$\Delta^2 = a^2 b^2 + 18 abc - 4 b^3 - 4 a^3 c - 27 c^2$$
The simplest way I know to prove this identity, is like this: let $x=x_1^2x_2+x_2^2x_3+x_3^2x_1$ and $y=x_1x_2^2+x_2x_3^2+x_3x_1^2$. Then:
$$\Delta^2=(x-y)^2=(x+y)^2-4xy$$
It's pretty simple to notice that $x+y=ab-3c$ and for $xy$, expanding:
$$xy=c(x_1^3+x_2^3+x_3^3)+(x_1^3x_2^3+x_2^3x_3^3+x_3^3x_1^3)+3c^2$$
and for the sum of cubes we have the well-known factorization:
$$x_1^3+x_2^3+x_3^3 = 3c+a(a^2-3b)$$
and similarly:
$$x_1^3x_2^3+x_2^3x_3^3+x_3^3x_1^3=3c^2+b(b^2-3ca)$$
Replacing back all of this:
$$
\begin{aligned}
xy &= c[3c+a(a^2-3b)]+[3c^2+b(b^2-3ca)]+3c^2\\
&= b^3 - 6 a b c + 9 c^2 + ca^3 
\end{aligned}
$$
and thus:
$$
\begin{aligned}
\Delta^2 &= (ab-3c)^2-4(b^3 - 6 a b c + 9 c^2 + ca^3 )\\
&= a^2b^2+18abc-4b^3-4a^3c-27c^2
\end{aligned}
$$
And we can determine $a,b,c$ from Vieta's ($a=2, b=3, c= -5$). In the end $\Delta^2=-1127$.
A: 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
$$(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2=$$
$$=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)=$$
$$=27\left(\frac{4}{3}-4+\frac{160}{27}-20-25\right)=-1127.$$
A: For a cubic equation $ax^3+bx^2+cx+d=0$,
the roots $x_1, x_2, x_3$ have the following properties:
$x_1+x_2+x_3=-\frac ba$
$x_1x_2+x_1x_3+x_2x_3=\frac ca$
$x_1x_2x_3=-\frac da$
Since you have $x^3-2x^2+3x+5=0$,
$x_1+x_2+x_3=2$
$x_1x_2+x_1x_3+x_2x_3=3$
$x_1x_2x_3=-5$
Can you do the rest?
A: The 'easy way' to do this leans on power sums instead of elementary symmetric polynomials.  We are implicitly working in $\mathbb C$ for this problem.  
For any degree $n$ monic polynomial, first encode it in an $n$ x $n$ Companion matrix $C$.   
now consider the power sum for $k\in\{1,2,3,...,\}$
$s_k := \lambda_1^k + \lambda_2^k +....+ \lambda_n^k = \text{trace}\big(C^k\big)$ and
$s_0:= n$
and $\lambda_i$ are the roots to your polynomial / the eigenvalues of $C$ 
now consider the matrix
$M_n := \begin{bmatrix}
s_0 & s_1  & s_2 & \cdots & s_{n-1}\\ 
s_1 &  s_2&  s_3 & \cdots & s_n \\ 
s_2& s_3 & s_4 & \cdots & s_{n+1}\\ 
\vdots & \vdots & \vdots & \ddots & \vdots\\ 
s_{n-1} & s_{n} & s_{n+1}  & \cdots  & s_{2n-2}
\end{bmatrix}$
for your problem here it is just  
$M_3 := \begin{bmatrix}
s_0 & s_1  & s_2\\ 
s_1 &  s_2&  s_3\\ 
s_2& s_3 & s_4 \\ 
\end{bmatrix}$
(note: matrix multiplication is only needed to get $s_2$.  You get $s_0$ and $s_1$ immediately.  Also $s_3$ and $s_4$ are obtainable by Cayley Hamilton.)  
and
$\det\big(M_3\big) = \Delta^2 = (\lambda_2-\lambda_1)^2(\lambda_3-\lambda_1)^2(\lambda_3-\lambda_2)^2$ 
because
$M= V^TV \longrightarrow \det\big(M\big)=\det\big(V^TV\big)=\det\big(V^T\big)\det\big(V\big)=\det\big(V\big)^2$
(note that is a transpose, not conjugate transpose.  This factorization turns out to be quite useful.)  
where, for avoidance of doubt, $V$ is the Vandermonde matrix, shown below for the $n=3$ case
$V := \begin{bmatrix}
1 & \lambda_1  & \lambda_1^2 \\ 
1 &  \lambda_2&  \lambda_2^2\\ 
1 & \lambda_3 & \lambda_3^2 \\ 
\end{bmatrix}$ 
