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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