Question

Alpha,beta,gamma are zeroes of polynomial kx^3+5x+9 . If alpha^3+beta^3+gamma^3=27. Find value of k.

Answer 1

$\alpha,\beta$ and $\gamma$ are zeroes of polynomial kx³ + 5x + 9.

so, $\alpha+\beta+\gamma=0$

$\alpha\beta+\beta\gamma+\gamma\alpha=\frac{5}{k}$

$\alpha\beta\gamma=-\frac{9}{k}$......(1)

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

= $0\times\{\alpha^2+\beta^2+\gamma^2-\frac{5}{k}\}$

= 0

so, $\alpha^3+\beta^3+\gamma^3-3\alpha\beta\gamma=0$

or, $\alpha^3+\beta^3+\gamma^3=3\alpha\beta\gamma$

but it is given that,

$\alpha^3+\beta^3+\gamma^3=27$

so, $3\alpha\beta\gamma=27$

or, $\alpha\beta\gamma=9$

from equation (1),

- 9/K = 9 => K = -1

Answer 2

Step-by-step explanation:

hi dear

kx³= a , 5x = b, 9= c

a+ b+ c = coefficient of x²/ coefficient of x³

coefficient of x²= 0

if a + b + c = 0

a³+ b³ + c ³= 3 abc

3abc= 27. ( a³ +b³ + c³= 27)

abc= 9

abc = constant term /cofficient of x ²

abc= -9/k

9= -9/ k

k= -1

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