the value of :$ ((a-b)(b-c)(c-a))^2$ If the polynomial : $f(x)=x^3-3x+2$ have the roots :$a,b,c$
How to find the value of :$$ ((a-b)(b-c)(c-a))^2$$
 A: We have $f'(x) = 3x^2 - 3$, so $f'(x) = 0$ when $x = \pm 1$. We also have $f(1) = 1^3 - 3(1) + 2 = 0$, so $1$ is in fact a double root. Can you now determine the value of the desired expression?
A: Notice that $f(x)=(x+2)(x-1)(x-1)$. This means the polynomial has three real roots which are $-2$, $1$ and $1$. Therefore you have that $((a-b)(b-c)(c-a))^2=(a-b)^2(b-c)^2(c-a)^2$. Notice that you have a cyclic product of squares. This means the product will be invariant of your choice of $a$, $b$ and $c$. You get $((a-b)(b-c)(c-a))^2=(-2-1)^2(1-1)^2(1-2)^2=0$. Finally, if you are wondering how to get the factoring of the polynomial, a good strategy is to guess a root and then if the root is $a\in\mathbb{R}$ divide the polynomial by $x-a$. The likelihood of guessing the root in little time will come with exposure to such polynomial ring exercises.
A: $1$ is a root of $f(x)$. Use long division to get the other roots. Finally, substitute for $a,b,c$
in $(a-b)^2(a-c)^2(b-c)^2$.
A: That is the discriminant of your polynomial. It's straightforward to calculate if you're comfortable with resultants: $\text{Disc}(f) = \text{Res}(f, f')$ when $f$ is monic.
