Question

If

$\alpha \: \beta \: \gamma \: are \: roots \: of \: {x}^{3} + {px}^{2} + qx + r = 0 \:$

form an equation whose roots are

$\alpha - \frac{1}{ \beta \gamma } , \beta - \frac{1}{ \gamma \alpha } , \gamma - \frac{1}{ \alpha \beta }$
​

Answer 1

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

Given that,

$\rm \: \alpha, \beta, \gamma \: are \: roots \: of \: {x}^{3} + {px}^{2} + qx + r = 0$

So,

$\rm\implies \: \alpha + \beta + \gamma = - \: \: p$

$\rm\implies \: \alpha \beta + \beta \gamma + \gamma \alpha = q$

and

$\rm\implies \: \alpha \beta \gamma \: = \: - \: r$

Let assume that the transformed equation is in y.

So, it means

$\rm \: y = \alpha - \dfrac{1}{ \beta \gamma }$

can be rewritten as

$\rm \: y = \alpha - \dfrac{ \alpha }{ \alpha \beta \gamma }$

$\rm \: y = \alpha - \dfrac{ \alpha }{ - r}$

$[\rm \: \because \: \alpha \beta \gamma = - r \: ]$

$\rm \: y = \alpha \bigg[1 + \dfrac{1}{r} \bigg]$

$\rm \: y = \alpha \bigg[\dfrac{r + 1}{r} \bigg]$

As it is given that,

$\rm \: \alpha, \beta, \gamma \: are \: roots \: of \: {x}^{3} + {px}^{2} + qx + r = 0$

$\rm\implies \: \alpha = x$

On substituting this value in above expression, we get

$\rm \: y = x\bigg[\dfrac{r + 1}{r} \bigg]$

$\rm\implies \:x = \dfrac{ry}{r + 1}$

So, on substituting this value of x in given polynomial

$\rm \: {x}^{3} + {px}^{2} + qx + r = 0$

we get,

$\rm \: {\bigg[\dfrac{yr}{r + 1} \bigg]}^{3} + p {\bigg[\dfrac{yr}{r + 1} \bigg]}^{2} + q\bigg[\dfrac{yr}{r + 1} \bigg] + r = 0$

On taking LCM, we get

$\rm \: {y}^{3} {r}^{3} + {pr}^{2} {y}^{2}(r + 1) + qry {(r + 1)}^{2} + r {(r + 1)}^{3} = 0$

On dividing whole equation by r, we get

$\rm \: {y}^{3} {r}^{2} + {pr} {y}^{2}(r + 1) + qy {(r + 1)}^{2} + {(r + 1)}^{3} = 0$

Hence, the required equation is

$\\ \rm \: {r}^{2} {y}^{3} + {pr}(r + 1) {y}^{2} + q {(r + 1)}^{2} y+ {(r + 1)}^{3} = 0$

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Formulae Used :-

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

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

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

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