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 :$$ ((a-b)(b-c)(c-a))^2$$

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.
