Question

if alpha beta gamma are the zeroes of the polynomial x3 + px2 + qx + r, then find 1/alpha beta + 1/beta gamma + 1/gamma alpha​

Answer 1

$\large\underline{\sf{Given- }}$

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: {x}^{3} + p{x}^{2} + qx + r\:}$

$\large\underline{\sf{To\:Find - }}$

$\boxed{ \rm{ \frac{1}{ \alpha \beta } + \frac{1}{ \beta \gamma } + \frac{1}{ \gamma \alpha }}}$

$\large\underline{\sf{Solution-}}$

We know that,

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then}$

$\purple{\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}}$

$\purple{\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}}}$

$\purple{\boxed{ \bf{ \: \alpha \beta \gamma = - \: \dfrac{d}{a}}}}$

Now, Given that,

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: {x}^{3} + p{x}^{2} + qx + r\:}$

$\rm :\longmapsto\:{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{p}{1} = - p}}$

and

$\rm :\longmapsto\:{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{r}{1} = - r}}$

Now, Consider,

$\rm :\longmapsto\:\dfrac{1}{ \alpha \beta } + \dfrac{1}{ \beta \gamma } + \dfrac{1}{ \gamma \alpha }$

$\rm \: = \: \: \dfrac{ \gamma + \alpha + \beta }{ \alpha \beta \gamma }$

$\rm \: = \: \: \dfrac{ \alpha + \beta + \gamma }{ \alpha \beta \gamma }$

$\rm \: = \: \: \dfrac{ - p}{ - r}$

$\rm \: = \: \: \dfrac{p}{r}$

Hence,

$\purple{\bf :\longmapsto\:\dfrac{1}{ \alpha \beta } + \dfrac{1}{ \beta \gamma } + \dfrac{1}{ \gamma \alpha } = \dfrac{p}{r} }$

Additional Information :-

$\red{\rm :\longmapsto\: \alpha , \beta \: are \: zeroes \: of \: a {x}^{2} + b {x}^{} + c, \: then}$

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

and

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$