Question

alpha beta gamma are zeros of a cubic polynomial kx cube minus 5 x + 9 if Alpha Cube + beta cube plus gamma cube is equal to 27 find the value of k

Answer 1

heya...

Here is your answer it may help....☺☺

Answer 2

Answer:

$\red {Value \: of \: k }\green {=-1}$

Step-by-step explanation:

$Compare \: kx^{3}-0\times x^{2}-5x+9\:with\\ax^{3}+bx^{2}+ cx +d,\: we \: get$

$a = k , \: b = 0,\: c = -5, \:d = 9$

$Given \: \alpha , \: \beta \: and \: \gamma \\ar \: zeroes \: of \: cubic \: polynomial$

$i)\blue { \alpha + \beta + \gamma} = \frac{-b}{a}$

$= \frac{-0}{k}}$

$\blue {= 0}\: ---(1)$

$ii)\pink { \alpha \beta \gamma} = \frac{-d}{a}$

$= \pink {\frac{-9}{k}}\:---(2)$

$\alpha^{3} + \beta^{3}+\gamma^{3} = 27 \:(Given)$

$\implies (\alpha + \beta + \gamma)[\alpha^{2} + \beta^{2} + \gamma^{2}-\alpha \beta - \beta \gamma - \gamma \alpha] + 3\alpha \beta \gamma = 27$

$\implies \blue {0} \times [\alpha^{2} + \beta^{2} + \gamma^{2}-\alpha \beta - \beta \gamma - \gamma \alpha] + 3\times \pink {\frac{-9}{k}} = 27$

$[ from \: (1) \: and \: (2) ]$

$\implies \frac{-27}{k} = 27$

$\implies \frac{-27}{27} = k$

$\implies \green { k = -1 }$

Therefore.,

$\red {Value \: of \: k }\green {=-1}$

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