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

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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – Julian Kuelshammer Jan 19 '13 at 12:50

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 '13 at 12:57
@dis You don't. – Thomas Andrews Jan 19 '13 at 13:02
@dis: Do you need to know which one is which? – Michael Albanese Jan 19 '13 at 13:21
If i don't need ,why ? – dis Jan 19 '13 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 '13 at 13:35

$1$ 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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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 '13 at 3:02