Question

if alpha beta gamma are the zeros of polynomial x³ + p X³ + q x + R then 1 by Alpha²+ 1 by beta²+ 1 by gamma ²​

Answer 1

ANSWER

q^2+q / R^2

EXPLANATION

see attachment

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Answer 2

$Given \: \: Question \: Is \: \\ \\ \alpha \: \: \: \beta \: \: \: and \: \: \: \gamma \: \: \: are \: \: \: the \: \: \: the \: \: \: zeroes \: \: of \: \: \\ f(x) = x {}^{3} + px {}^{2} + qx + r \\ find \: \: \: \frac{1}{ \alpha {}^{2} } + \frac{1}{ \beta {}^{2} } + \frac{1}{ \gamma {}^{2} } \\ \\ Answer \: \\ \\ \frac{1}{ \alpha {}^{2} } + \frac{1}{ \beta {}^{2} } + \frac{1}{ \gamma {}^{2} } = \frac{( \alpha {}^{2} + \beta {}^{2} + \gamma {}^{2} }{( \alpha {}^{2} \beta {}^{2} \gamma {}^{2} ) } \\ \\ for \: \: a \: \: general \: \: cubic \: \: polynomial \: \: \: say \: \: \\ g(x) = ax {}^{3} + b {}^{3} + cx + d \\ we \: \: have \: \\ \\ ( \alpha + \beta + \gamma ) = \frac{ - b}{a} \\ ( \alpha \beta + \beta \gamma + \alpha \gamma ) = \frac{c}{a} \\ and \: \: \: \alpha \beta \gamma = \frac{ - d}{a} \\ \\ here \\ \\ ( \alpha + \beta + \gamma ) = - p \\ ( \alpha \beta + \beta \gamma + \alpha \gamma ) = q \\ and \\ ( \alpha \beta \gamma ) = - r \\ \\let \: \: \: x = \frac{( \alpha {}^{2} + \beta {}^{2} + \gamma {}^{2} )}{( \alpha {}^{2} \beta {}^{2} \gamma {}^{2} ) } = \frac{( \alpha + \beta + \gamma ) {}^{2} - 2( \alpha \beta + \beta \gamma + \alpha \gamma )}{( \alpha {}^{2} \beta {}^{2} \gamma {}^{2} )} \\ \\ x = \frac{( - p {}^{2} - 2q) }{r {}^{2} } \\ \\ therefore \: \: \: \\ \\ \frac{1}{ \alpha {}^{2} } + \frac{1}{ \beta {}^{2} } + \frac{1}{ \gamma {}^{2} } = \frac{ - (p {}^{2} + 2q) }{r {}^{2} } \\ \\ note \: \: \\ \\ (x + y + z) {}^{2} = x {}^{2} + y {}^{2} + z {}^{2} + 2(xy + yz + zx)$