Question

1. Alpha, beta, gemma are zeroes of cubic polynomial kx^3-5x+9. If alpha^3+beta^3+gemma^3= 27, find the value of k.

Answer 1

Heya. !

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Given cubic polynomial is:  kx^3 - 5x + 9 = 0.

It can be rewritten as:

 kx^3 + 0x^2 - 5x + 9 = 0

It is given that a,b and c are the roots/zeros of the cubic polynomial such that a^3 + b^3 + c^3 = 27  ...(1)

Now, 

sum of roots = a+b+c = -0/k = 0

sum of product of roots taken two at a time = ab+ bc +ca = -5/k

Product of roots = abc = -9/k .... (2)

We know that if, a+b+c = 0

then a^3 + b^3 + c^3 = 3abc ..... (3)

On equating (1) and (3), we get

3abc = 27

3(-9/k) = 27  {using (2)}

⇒ -27/k= 27

⇒ k = -1

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Hope it helps u !!!!

Cheers :))

# Nikky