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Let $a$, $b$, $c$ be the roots of $$x^3 - 6x^2 + 3x + 1 = 0$$ Find all the possible values of $a^2b + b^2c + c^2a$ and hence find $|(a-b)(b-c)(c-a)|$.

My try :

Let $P(x) = x^3-6x^2+3x+1 = (x-a)(x-b)(x-c)$
Then, $\frac{P(x)}{x-a} = (x-b)(x-c) => (a-b)(a-c) = \lim{x → a} \frac{x^3-6x^2+3x+1}{x-a} = 3(a^2-4a+1) = 3(a-(2+√3))(a-(2-√3))$.

Similarly, $(b-a)(b-c) = 3(b^2-4b+1) = 3(b-(2+√3))(b-(2-√3))$ and $(c-a)(c-b) = 3(c^2-4c+1) = 3(c-(2+√3))(c-(2-√3))$ .

Combining them, we get $(a-b)^2(b-c)^2(c-a)^2 = 27P(2+√3)P(2-√3)$.

How to proceed from here ??

Ash_Blanc
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    $((a - b) (b - c) (c - a))^2 = -27 a^2 b^2 c^2 - 4 a b c (a + b + c)^3 + 18 a b c (a + b + c) (a b + a c + b c) + (a + b + c)^2 (a b + a c + b c)^2 - 4 (a b + a c + b c)^3$ – Basics Jun 12 '24 at 02:18
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    @Basics Add a sentence or two and put this as an answer. (Please don't answer questions in the comments as the question stays open indefinitely this way.) – Sammy Black Jun 12 '24 at 02:22
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    meanwhile, $x^3 - 6 x^2 + 3x + 1$ has the same discriminant as $t^3 - 9t - 9$ which is a pure shift, take $x = t + 2.$ And your number is the square root of the discriminant, which is positive and an evident square – Will Jagy Jun 12 '24 at 02:31
  • and the discriminant of $t^3 + ut + v$ is $-4 u^3 - 27 v^2 $ – Will Jagy Jun 12 '24 at 02:38
  • By depressed cubic $P(2+\sqrt3)P(2-\sqrt3)=-27$. Then my answer has $i$ mistake? – Bob Dobbs Jun 12 '24 at 07:30

2 Answers2

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By what you did, you know that $$ |(a-b)(b-c)(c-a)|=|(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)|=\sqrt{27P(2+\sqrt{3})P(2-\sqrt{3})}=27, $$ so we know the difference of $ab^2+bc^2+ca^2$ and $a^2b+b^2c+c^2a$. Let us try to find their sum!

We have $$ (a+b+c)(ab+bc+ca)=3abc+\sum_{\rm sym}a^2b. $$ From the coefficients of $P$, we know that $a+b+c=6$, $ab+bc+ca=3$ and $abc=-1$. Hence we obtain $$ \sum_{\rm sym}a^2b=3+3\cdot 6=21. $$ Therefore, the numbers $ab^2+bc^2+ca^2$ and $a^2b+b^2c+c^2a$ have difference $27$ and sum $21$. Hence we obtain $a^2b+b^2c+c^2a\in\{-3,\ 24\}$. It is clear that both are possible, because if we replace pose $(\alpha,\beta,\gamma)=(a,c,b)$, then we still have $P(x)=(x-\alpha)(x-\beta)(x-\gamma)$, and $$ \alpha^2\beta+\beta^2\gamma+\gamma^2\alpha=ab^2+bc^2+ca^2. $$ Hence if $(a,b,c)$ yields an element of $\{-3,\ 24\}$, then $(\alpha,\beta,\gamma)$ satisfies the same hypotheses and yields the other element.

imtrying46
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Hint: By Cardano's formula, the simplification is $Z=6\sqrt{3}|i\sqrt{\delta}|$ where $\delta$ is the discriminant of the depressed cubic.

Let $x=y-\frac{-6}{3}=y+2$. Then OP's cubic transforms to $y^3-9y-9=0$. Its discriminant is $\delta=\frac{(-9)^2}{4}+\frac{(-9)^3}{27}=-\frac{27}4$. And $Z=6\sqrt3|i^2\frac{3\sqrt3}2|=27.$

Bob Dobbs
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