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February 16, 2021

Test of Mathematics Solution Subjective 72 - Polynomial Problem

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 72 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


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Problem

If $ {\displaystyle{\alpha}, {\beta}, {\gamma}} $ are the roots of the equation $ {\displaystyle{x^3 + px^2 + qx + r = 0}} $, find the equation whose roots are $ {\displaystyle{\alpha} - {\frac{1}{{\beta}{\gamma}}}} $, $ {\displaystyle{\beta} - {\frac{1}{{\alpha}{\gamma}}}} $, $ {\displaystyle{\gamma} - {\frac{1}{{\alpha}{\beta}}}} $ .


Solution


${\displaystyle{\alpha}, {\beta}, {\gamma}} $ are roots of $ {\displaystyle{x^3 + px^2 + qx + r = 0}} $
${\Rightarrow} $ $ {\displaystyle{\alpha}, {\beta}, {\gamma}} $ = $ {- r} $,
$ {\alpha + \beta + \gamma} $ = $ {- p} $
$ {{\alpha}{\beta} + {\gamma}{\alpha} + {\beta}{\gamma}} $ = $ {q} $
Now,
$ {\displaystyle{\left({\alpha} - {\frac{1}{{\beta}{\gamma}}}\right)}} $ + $ {\displaystyle{\left({\beta} - {\frac{1}{{\alpha}{\gamma}}}\right)}} $ + $ {\displaystyle{\left({\gamma} - {\frac{1}{{\alpha}{\beta}}}\right)}} $

= $ {\displaystyle{\frac{{\alpha}{\beta}{\gamma} - 1}{{\alpha}{\beta}{\gamma}}}} $ $ {\displaystyle{\left({\alpha} + {\beta} + {\gamma}\right)}} $

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

$ {\sum}$ $ {\displaystyle{\left({\alpha} - {\frac{1}{{\beta}{\gamma}}}\right)}} $ $ {\displaystyle{\left({\beta} - {\frac{1}{{\alpha}{\gamma}}}\right)}} $

= $ {\sum} $ $ {\displaystyle{\alpha \beta} - 2{\frac{1}{\gamma} + {\frac{1}{\alpha \beta \gamma^2}}}} $

= $ {\displaystyle{(\alpha \beta + \beta \gamma + \gamma \alpha)}} $ - 2$ {\displaystyle{\left({\frac{1}{\alpha}} + {\frac{1}{\beta}} + {\frac{1}{\gamma}}\right)}} $ + $ {\displaystyle{\frac{1}{\alpha \beta \gamma}}} $ + $ {\displaystyle{\left({\frac{1}{\alpha}}+{\frac{1}{\beta}} + {\frac{1}{\gamma}}\right)}} $

= $ {\displaystyle{q} - 2 {\left({\frac{q}{-r}}\right)} + {\frac{1}{-r}} {\left({\frac{q}{-r}}\right)}} $

= $ {\displaystyle{q}\left(1 + {\frac{1}{r}}\right)^2} $

$ {\displaystyle{\left({\alpha} - {\frac{1}{\beta \gamma}}\right)}} $ + $ {\displaystyle{\left({\beta} - {\frac{1}{\alpha \gamma}}\right)}} $ + $ {\displaystyle{\left({\gamma} - {\frac{1}{\alpha \beta}}\right)}} $

= $ {\displaystyle{\alpha \beta \gamma}} $ - 3 + $ {\displaystyle{3}{\frac{1}{\alpha \beta \gamma}}} $ - $ {\displaystyle \left({\frac{1}{\alpha \beta \gamma}}\right)^2} $

= - $ {\displaystyle \left(r + 3 + 3 {\frac{1}{r}} + {\frac{1}{r^2}}\right)} $

So the equation whose roots are ${\displaystyle{\alpha} - {\frac{1}{{\beta}{\gamma}}}} $, $ {\displaystyle{\beta} - {\frac{1}{{\alpha}{\gamma}}}} $, $ {\displaystyle{\gamma} - {\frac{1}{{\alpha}{\beta}}}} $ is ${\displaystyle{x^3} + {\frac{p(r+1)}{r}} x^2 + q \left(1+{\frac{1}{r}}\right)^2{x} + \left(r+3+{\frac{3}{r}}+{\frac{1}{r^2}}\right)} $

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