Given that $x_1, x_2, x_3$ are the roots of the polynomial $x^3-2x^2+3x+5=0$ find $(x_2-x_1)^2(x_3-x_1)^2(x_3-x_2)^2$.

$\Delta$ 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 $\Delta^2=-1127$.

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 $$(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2=$$ $$=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)=$$ $$=27\left(\frac{4}{3}-4+\frac{160}{27}-20-25\right)=-1127.$$

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?