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Polynomial Problem (Tomato subjective 72)

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 \({\displaystyle{x^3} + {\frac{p(r+1)}{r}} x^2 + q \left(1+{\frac{1}{r}}\right)^2{x} + \left(r+3+{\frac{3}{r}}+{\frac{1}{r^2}}\right)} \)

August 22, 2015
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