# Complex root term

This is a question from a live contest site and therefore I am asking for clarification, please do not answer the actual question. Thanks.

The question:

Question removed.

I have written what I have done here and the document will last only 30 minutes so as not to be seen unwittingly in times to come: link removed for preventative purposes.

Please can someone show me where I went wrong, but again please do not answer the question.

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Have you checked that you are allowed to solicit external help? Showing you where you went wrong is rather non-trivial interference with the contest. –  Alex B. Jan 1 '13 at 12:04
@fosho, I think there is a better approach by replacing $x$ with $\frac{1-y}{1+y}$, which I had adapted in my deleted answer. –  lab bhattacharjee Jan 1 '13 at 12:18
@Alex B, im not sure but I have really tried for a long time on this one and have made all 3 attempts and therefore cannot get the points for this question. –  fosho Jan 1 '13 at 12:34
@fosho If you did a Request Clarification and gave the answer of 5/9, then your numerator should have been $\textbf{3}+p+q+r -pq-pr-qr -3pq$, instead of (starting with) 1. This would give you $\frac {19}{9}$. Otherwise, ignore this comment. –  Calvin Lin Jan 6 '13 at 2:28

You could go hard on basic algebra and using Viete's relations:

$$\frac{1−p}{1+p}+\frac{1−q}{1+q}+\frac{1−r}{1+r}=$$

$$\frac{(1-p)(1+q)(1+r)+(1-q)(1+p)(1+r)+(1-r)(1+p)(1+q)}{(1+p)(1+q)(1+r)}$$

Now remember that $\,pqr=1/7\;\;,\;\;pq+pr+rq=0\,$ and etc.

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are you sure pqr = 1/7? –  fosho Jan 1 '13 at 12:48
@fosho : $$7x^3-x^2-1=7(x-p)(x-q)(x-r)\Longrightarrow -7pqr=-1\ldots$$ –  DonAntonio Jan 1 '13 at 12:50
Thanks! could you please remove you answer so as not to be unwittingly found by people still solving this problem. Much Appreciated. –  fosho Jan 1 '13 at 12:55
Don't worry: without the actual question in your OP and the slight and fair, imo, hint given, odds are nobody will unfairly profit from it. –  DonAntonio Jan 1 '13 at 12:57
@DonAntonio I would appreciate if you could take this answer down. We do ask students to submit solutions, and I do not want to read several copies of yours. If you have any questions, please reply/email me. Thanks - Calvin Lin, Brilliant Challenge Master. –  Calvin Lin Jan 6 '13 at 2:25