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I tried using Vieta's relation and wanted to somehow generate a term which would cancel out using $abc=-1$ but was unsuccessful..

DominikS
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Rexquiem
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    Try Vieta's relation again. – uniquesolution Dec 13 '23 at 10:10
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    I you know about discriminants, use that knowledge. If not, take the square of that value and use Vieta's relations on that (the square is symmetric in $a,b,c$). – user8268 Dec 13 '23 at 10:13
  • @uniquesolution Using the Vieta's relations is not enough, since the expression is not symmetric in $a$, $b$, $c$. – jjagmath Dec 13 '23 at 18:30
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    One can only determine the value up to sign, since permuting the labels of the roots, e.g., by $a \leftrightarrow b$ reverses the sign. On the other hand the square of the expression is just the discriminant of the polynomial, which is invariant under permutations. – Travis Willse Dec 13 '23 at 19:03

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Let $$ P = (a-b)(b-c)(c-a) $$ Notice $$ (a-b)^3 = a^3 - b^3 - 3ab(a-b) $$ and $$ a^3 + 3a + 1 = 0 $$ $$ b^3 + 3b + 1 = 0 $$ so that $$ a^3 - b^3 = -3(a-b) $$ Then we have $$ (a-b)^3 = -3(1+ab)(a-b) $$ Then $$ P^3 = -27(1+ab)(1+bc)(1+ac)P $$ $$ P^2 = -27(1 + ac + bc + ab + b^2ac + a^2bc + c^2ab + a^2b^2c^2) = -27(1 + 3 + abc(a+b+c) + 1) = -135 $$ $$ P = \pm\sqrt{135}i $$ Depending on the actual values of $a,b,c$, the result is either $\sqrt{135}i$ or $-\sqrt{135}i$.

poeplva
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