Question

Alpha,beta,gamma are the zero of cubic polynomial kx^3-5x+9.if alpha^3+beta^3+gamma^3=27.find the value of k

Answer 1

Hey!

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Let, alpha, beta , gamma be a , b , c
[Will be easier for me to type ]

Given,

a , b , c are zeroes of kx^3 - 5x + 9

a^3 + b^3 + c^3 = 27

ka^3 -5a + 9 = 0
kb^3 -5a + 9 = 0
kc^3 -5a + 9 = 0

Add all three equations -

ka^3 - 5a+ 9 + kb^3 - 5a+9 + kc^3 -5a +9 = 0

k(a^3 + b^3 + c^3) - 5 (a+b+c) + 27 = 0 (i)

kx^3 - 5x + 9 has no x^2, thus a+b+c = 0 (ii)

Now, from (i) and (ii) ,

27k + 27 = 0
k = -1


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Hope it helps...!!!

Answer 2

The answer is given in the attachment.

We have found that, k = -1

Thank you for the question.