Difference between revisions of "2022 SSMO Team Round Problems/Problem 10"

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==Problem==
 
==Problem==
If <math>p, q, r</math> are the roots of the polynomial <math>x^3-2x^2-4</math>, find<cmath>(p^3+qr)(q^3+pr)(r^3+pq).</cmath>
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If <math>\alpha, \beta, \gamma</math> are the roots of the polynomial <math>x^3-2x^2-4</math>, find <cmath>(\alpha^3+\beta\gamma)(\beta^3+\gamma\alpha)(\gamma^3+\alpha\beta).</cmath>
  
 
==Solution==
 
==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>
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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>
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<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 11:32, 25 December 2023

Problem

If $\alpha, \beta, \gamma$ are the roots of the polynomial $x^3-2x^2-4$, find \[(\alpha^3+\beta\gamma)(\beta^3+\gamma\alpha)(\gamma^3+\alpha\beta).\]

Solution

By Vieta's relation we get, $\sum_{cyc}{}\alpha=2$ $\sum_{cyc}{}\alpha\beta=0$ and $\prod_{cyc}{}\alpha=4$

Therefore we have to find the value of \[\prod_{cyc}{}(\alpha^3+\beta\gamma)\implies\prod_{cyc}{}\left(\alpha^3+\frac{4}{\alpha}\right)\implies\prod_{cyc}{}\left(\frac{\alpha^4+4}{\alpha}\right)\] \[\prod_{cyc}{}\left(\frac{\alpha^4+4}{\alpha}\right)=\frac{\prod_{cyc}{}(\alpha-(1+i))(\alpha-(1-i))(\alpha+(1-i))(\alpha+(1+i))}{\alpha\beta\gamma}\]