Question

If $\mathsf{\alpha, \beta, \gamma}$ are the roots of $\mathsf{{x}^{3}+p{x}^{2}+qx+r = 0}$ the form the monic cubic equation whos roots are $\mathsf{\alpha ( \beta + \gamma ), \: \beta ( \gamma + \alpha ), \: \gamma( \alpha + \beta)}$

Answer 1

Since $\mathsf{\alpha , \beta , \gamma}$ be the roots of given equation, we have

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

$\alpha \beta + \beta \gamma + \gamma\alpha = q$

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

Let

$\sum\alpha( \beta + \gamma) \: = s_1$

$\sum \: \alpha \beta( \beta + \gamma)( \gamma + \alpha ) = s_2$

$and \: \: \alpha \beta \gamma( \beta + \gamma)( \gamma + \alpha)( \alpha + \beta) = s_ 3$

Then Required equation is $\mathsf{ {x}^{3} - S_1 {x}^{2} + S_2 x - S_3}$

Now we find $\mathsf{ s_1 , s_2 \: and \: s_3 }$ in terms of p,q and r.

$s_1 \: = \: \sum\alpha( \beta + \gamma) \: = \sum( \alpha \beta + \alpha \gamma)$

$\: = \sum \alpha \beta + \sum \alpha \gamma$

$= \sum \alpha \beta \: + \: \sum \alpha \beta$

$= 2\sum \alpha \beta = 2q$

_________

$s_2 = \: \sum \: \alpha \beta( \beta + \gamma)( \gamma + \alpha )$

$= \sum \alpha \beta( - p - \alpha)( - p - \beta)$

$= \sum \alpha \beta \Big \{ {p}^{2} + ( \alpha + \beta)p + \alpha \beta \Big \}$

$= \sum \alpha \beta\Big \{ {p}^{2} - (p + \gamma)p + \alpha \beta \Big \}$

$= \sum \alpha \beta \Big \{ - \gamma p + \alpha \beta \Big \}$

$= \sum { (\alpha}^{2} { \beta}^{2} - \alpha \beta \gamma p) = \sum { \alpha}^{2} { \beta}^{2} - p \sum \alpha \beta \gamma$

$= \sum { \alpha}^{2} { \beta}^{2} - p( - 3r)$

$= {( \alpha \beta + \beta \gamma + \gamma \alpha) }^{2} - 2 \alpha \beta \gamma( \alpha + \beta + \gamma) + 3pr$

$= {q}^{2} - 2pr + 3pr$

$= {q}^{2} + pr$

___________

$s_ 3 = \alpha \beta \gamma( \beta + \gamma)( \gamma + \alpha)( \alpha + \beta)$

$= - r( - p - \gamma)( - p - \alpha)( - p - \beta)$

$= r(p + \gamma)(p + \alpha)(p + \beta)$

$= r[ {p}^{3} + ( \alpha + \beta + \gamma) {p}^{2} + ( \alpha \beta + \beta \gamma + \gamma \alpha)p \: + \alpha \beta \gamma ]$

$= r[ {p}^{3} - {p}^{3} + pq - r ]$

$= r(pq - r)$

Therefore the required equation is :-

${x}^{3} - 2q {x}^{2} + ( {q}^{2} + pr)x - r(pq - r) = 0$

Answer 2

If $\mathsf{\alpha, \beta, \gamma}$ are the roots of $\mathsf{{x}^{3}+p{x}^{2}+qx+r = 0}$ the form the monic cubic equation whos roots are $\mathsf{\alpha ( \beta + \gamma ), \: \beta ( \gamma + \alpha ), \: \gamma( \alpha + \beta)}$

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