Difference between revisions of "2022 SSMO Team Round Problems/Problem 10"
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==Solution== | ==Solution== | ||
| − | By Vieta's relation we get, <math>\sum_{cyc}{}\alpha=2</math> <math>\sum_{cyc}{}\alpha\beta=0</math> and <math>\prod_{cyc}{}\alpha=4 | + | By Vieta's relation we get, <math>\sum_{cyc}{}\alpha=2</math> <math>\sum_{cyc}{}\alpha\beta=0</math> and <math>\prod_{cyc}{}\alpha=4</math> |
Therefore we have to find the value of <cmath>\prod_{cyc}{}(\alpha^3+\beta\gamma)=\prod_{cyc}{}\left(\alpha^3+\frac{4}{\alpha}\right)=\prod_{cyc}{}\left(\frac{\alpha^4+4}{\alpha}\right)</cmath> | Therefore we have to find the value of <cmath>\prod_{cyc}{}(\alpha^3+\beta\gamma)=\prod_{cyc}{}\left(\alpha^3+\frac{4}{\alpha}\right)=\prod_{cyc}{}\left(\frac{\alpha^4+4}{\alpha}\right)</cmath> | ||
<cmath>\prod_{cyc}{}\left(\frac{\alpha^4+4}{\alpha}\right)=\frac{\prod_{cyc}{}(\alpha-(1+i))(\alpha-(1-i))(\alpha+(1-i))(\alpha+(1+i))}{\alpha\beta\gamma}</cmath> | <cmath>\prod_{cyc}{}\left(\frac{\alpha^4+4}{\alpha}\right)=\frac{\prod_{cyc}{}(\alpha-(1+i))(\alpha-(1-i))(\alpha+(1-i))(\alpha+(1+i))}{\alpha\beta\gamma}</cmath> | ||
Revision as of 10:32, 25 December 2023
Problem
If 
are the roots of the polynomial 
, find ![\[(\alpha^3+\beta\gamma)(\beta^3+\gamma\alpha)(\gamma^3+\alpha\beta).\]](http://latex.artofproblemsolving.com/a/4/6/a463ccc3de2f3710c58da87d8e3465af873f82ad.png)
Solution
By Vieta's relation we get, 
and 
Therefore we have to find the value of ![\[\prod_{cyc}{}(\alpha^3+\beta\gamma)=\prod_{cyc}{}\left(\alpha^3+\frac{4}{\alpha}\right)=\prod_{cyc}{}\left(\frac{\alpha^4+4}{\alpha}\right)\]](http://latex.artofproblemsolving.com/b/c/5/bc54153c8c6153234ad5329797dad0f8ad45a8ce.png)
![\[\prod_{cyc}{}\left(\frac{\alpha^4+4}{\alpha}\right)=\frac{\prod_{cyc}{}(\alpha-(1+i))(\alpha-(1-i))(\alpha+(1-i))(\alpha+(1+i))}{\alpha\beta\gamma}\]](http://latex.artofproblemsolving.com/4/e/c/4ec33564055263744cec63501763130a8477ac0f.png)
