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If the polynomial : $f(x)=x^3-3x+2$ have the roots :$a,b,c$

How to find the value of :$$ ((a-b)(b-c)(c-a))^2$$

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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?

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But how can i know which one is $a$ and $b$ and $c$? – dis Jan 19 at 12:57
@dis You don't. – Thomas Andrews Jan 19 at 13:02
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@dis: Do you need to know which one is which? – Michael Albanese Jan 19 at 13:21
If i don't need ,why ? – dis Jan 19 at 13:28
@dis: It is worth trying the three possibilities for yourself. What is the result if you let $a$ and $b$ be $1$? What about if you used $a$ and $c$ instead? What about $b$ and $c$? More generally, if you swap any two of the letters in the expression, it doesn't change. For example, replace $a$ by $b$ and $b$ by $a$ then $((a-b)(b-c)(c-a))^2$ becomes $$\begin{align*}((b-a)(a-c)(c-b))^2 &= ((-1)(a-b)(a-c)(-1)(b-c))^2\\ &= ((a-b)(a-c)(b-c))^2\\ &= ((a-b)(b-c)(a-c))^2.\end{align*}$$ The same is true if you swap $a$ and $c$, or $b$ and $c$. – Michael Albanese Jan 19 at 13:35
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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.

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What If the polynomial is : $f(x)=x^3−3x+1$ rather than $f(x)=x^3−3x+2$? – dis Jan 20 at 3:02
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$1$ is a root of $f(x)$. Use long division to get the other roots. Finally, substitute for $a,b,c$
in $(a-b)^2(a-c)^2(b-c)^2$.

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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.

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