Difference between revisions of "2022 SSMO Team Round Problems/Problem 10"
Rounak iitr (talk | contribs) (→Solution) |
Rounak iitr (talk | contribs) (→Solution) |
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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> | 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) | + | Therefore we have to find the value of <cmath>\prod_{cyc}{}(\alpha^3+\beta\gamma)\implies\prod_{cyc}{}\left(\alpha^3+\frac{4}{\alpha}\right)\implies\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> |
Latest revision as of 10:32, 25 December 2023
Problem
If are the roots of the polynomial , find
Solution
By Vieta's relation we get, and
Therefore we have to find the value of