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