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)}