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*It is a question unanswered in my earlier post at: https://math.stackexchange.com/a/2755962/424260. *

I have been given that: $(y_1-y_2)(y_2-y_3)(y_3-y_1) = y_1y_2(y_1-y_2)+y_2y_3(y_2-y_3)+y_1y_3(y_3-y_1)$.

But, the derivation below shows that there is reversal of terms inside each of the $3$ brackets.
$$(y_1-y_2)(y_2-y_3)(y_3-y_1) = (y_1y_2-y_1y_3-y_2^2+y_2y_3)(y_3-y_1)$$ $$=-y_1y_3^2 -y_2^2y_3 +y_2y_3^2-y_1^2y_2+y_1^2y_3+y_1y_2^2$$ $$=y_1y_3(y_1-y_3)+y_2y_3(y_3-y_2)+y_1y_2(y_2-y_1)$$

Kindly help me with finding error in my derivation.

jiten
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  • So what's your question? Your given that $$(y_1-y_2)(y_2-y_3)(y_3-y_1) = -(\cdots) = (\cdots),$$ so the product is zero. – GNUSupporter 8964民主女神 地下教會 Apr 27 '18 at 11:29
  • @GNUSupporter Cannot understand, it is a question unanswered in my earlier post of today at: https://math.stackexchange.com/a/2755962/424260. You seem to say that the product of $(y_2-y_1)(y_3-y_2)(y_1-y_3)$ will yield the desired answer & hence the product is zero. It is not logical in context of my earlier post. – jiten Apr 27 '18 at 11:36
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    You're given that $$(y_1-y_2)(y_2-y_3)(y_3-y_1) = y_1y_2(y_1-y_2)+y_2y_3(y_2-y_3)+y_1y_3(y_3-y_1) \tag{1}.$$ $$y_1y_2(y_1-y_2)+y_2y_3(y_2-y_3)+y_1y_3(y_3-y_1) = -[y_1y_3(y_1-y_3)+y_2y_3(y_3-y_2)+y_1y_2(y_2-y_1)] \tag{2}$$ Expanding $(1)$ gives $$(y_1-y_2)(y_2-y_3)(y_3-y_1) = y_1y_3(y_1-y_3)+y_2y_3(y_3-y_2)+y_1y_2(y_2-y_1).\tag{3}$$ Conclusion: $$ y_1y_3(y_1-y_3)+y_2y_3(y_3-y_2)+y_1y_2(y_2-y_1) = -[ y_1y_3(y_1-y_3)+y_2y_3(y_3-y_2)+y_1y_2(y_2-y_1)] = 0$$ – GNUSupporter 8964民主女神 地下教會 Apr 27 '18 at 11:41
  • @GNUSupporter Please note that this is having a non-zero solution as shown at the top of the edited OP. So, please help by telling why the earlier post (of which it is unanswered question) is having this fallacy. – jiten Apr 27 '18 at 11:45
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    Sorry, I don't understand your logic. Unless you [edit] your post, mentioning another linked post is red herring. If something given in a math question is false and can't be used, I don't think we can do anything. – GNUSupporter 8964民主女神 地下教會 Apr 27 '18 at 11:49
  • @GNUSupporter How to edit my post more than stating the link to the earlier post, which has links to all the relevant pages of the book, and my derivaton towards that end, leading to the above conclusion. – jiten Apr 27 '18 at 11:51
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    The logical conclusion of the original version of the question is $(y_1-y_2)(y_2-y_3)(y_3-y_1) = 0$, so $y_1 = y_2$ or ... Both $y_i = 0$ or $y_i \ne 0$ are possible, so the possibility of having nonzero solution is not excluded. I meant that you edit to change the given conditions or the desired conclusion in the original version of your question when I typed my previous comment. However, since you've already received an answer, I will no longer suggest it. – GNUSupporter 8964民主女神 地下教會 Apr 27 '18 at 12:04

2 Answers2

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The given statement $$(y_1-y_2)(y_2-y_3)(y_3-y_1) = y_1y_2(y_1-y_2)+y_2y_3(y_2-y_3)+y_1y_3(y_3-y_1)$$ is not correct. Just check it for $$ y_1=1, y_2=2, y_3 =3$$ and you get $2=-2$

The correct version is $$(y_1-y_2)(y_2-y_3)(y_3-y_1)=y_1y_3(y_1-y_3)+y_2y_3(y_3-y_2)+y_1y_2(y_2-y_1)$$ which you have obtained.

Mohammad Riazi-Kermani
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  • But it is a valid problem with reference to my earlier post on finding solution to cubic general equation, as shown at the top of the edited OP. – jiten Apr 27 '18 at 11:46
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    It is valid for special case where both sides are zero, otherwise we have a problem with a negative sign. – Mohammad Riazi-Kermani Apr 27 '18 at 12:03
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The statement you have been given is missing a minus sign on one end.

Your deductions are absolutely correct. Notice that:

$$y_1y_2(y_1-y_2)=-y_1y_2(y_2-y_1)$$ and the same holds with the other two sets.

Rhys Hughes
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  • But it is a valid problem with reference to my earlier post on finding solution to cubic general equation, as shown at the top of the edited OP. – jiten Apr 27 '18 at 11:43
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    Where you've used this, to make it work you need to change it to: $$(y_1-y_2)(y_2-y_3)(y_3-y_1) = -[y_1y_2(y_1-y_2)+y_2y_3(y_2-y_3)+y_1y_3(y_3-y_1)]$$ – Rhys Hughes Apr 27 '18 at 11:51
  • Do you mean that the book has a sign error on page 3? – jiten Apr 27 '18 at 12:15
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    Yes, most likely it does. – Rhys Hughes Apr 27 '18 at 12:21
  • But, why the error does not affect the solution in the page $4$ then. It seems that sign change is immaterial. Also, the book refers on the beginning of pg. $4$, about usage of 'factor theorem', which I hope must be a reference to the 'reverse' derivation of equality $D$ by the product of three terms $(y_2-y_1)(y_3-y_2)(y_1-y_3)$. – jiten Apr 27 '18 at 12:35